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.github/workflows/typst.yml vendored Normal file
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name: Build Typst document
on: [push, workflow_dispatch]
permissions:
contents: write
jobs:
build:
runs-on: ubuntu-latest
steps:
- name: Checkout
uses: actions/checkout@v3
- name: Typst
uses: lvignoli/typst-action@main
with:
source_file: |
main.typ
- name: Upload PDF file
uses: actions/upload-artifact@v3
with:
name: PDF
path: "*.pdf"
- name: Get current date
id: date
run: echo "DATE=$(date +%Y-%m-%d-%H:%M)" >> $GITHUB_ENV
- name: Release
uses: softprops/action-gh-release@v1
if: github.ref_type == 'tag'
with:
name: "${{ github.ref_name }} — ${{ env.DATE }}"
files: main.pdf

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main.pdf *.pdf

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MIT License
Copyright (c) 2024 Kristofers Solo
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
SOFTWARE.

BIN
export/MathLogicCheatsheet.pdf
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main.typ
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@ -1,25 +1,12 @@
#set page( #set page(margin: 1cm, columns: 2)
margin: (
left: 1cm,
right: 1cm,
top: 1cm,
bottom: 1cm,
),
)
#set heading(numbering: "1.") #set heading(numbering: "1.")
#show outline.entry.where(level: 1): it => {
#show outline.entry.where(
level: 1
): it => {
v(12pt, weak: true) v(12pt, weak: true)
strong(it) strong(it)
} }
#set enum(numbering: "a)") #set enum(numbering: "a)")
// #show outline.entry.where( // #show outline.entry.where(
@ -31,14 +18,11 @@
#outline(indent: auto) #outline(indent: auto)
= Logical axiom schemes = Logical axiom schemes
$bold(L_1): B→(C →B)$ $bold(L_1): B→(C →B)$
$bold(L_2): (B→(C →D))→((B→C)→( B→D))$ $bold(L_2): (B→(C →D))→((B→C)→( B→D))$
$bold(L_3): B∧C→B$ $bold(L_3): B∧C→B$
@ -58,9 +42,9 @@ $bold(L_10): ¬B→( B→C)$
$bold(L_11): B∨¬B$ $bold(L_11): B∨¬B$
$bold(L_12): ∀x F (x)→F (t)$ (in particular, $∀x F (x)→F (x)$) $bold(L_12): ∀x F (x)→F (t)$ (in particular, $∀x F (x)→F (x)$)
$bold(L_13): F (t)→∃ x F( x)$ (in particular, $F (x)→∃ x F (x)$) $bold(L_13): F (t)→∃ x F( x)$ (in particular, $F (x)→∃ x F (x)$)
$bold(L_14): ∀x(G →F (x))→(G→∀x F (x)) $ $bold(L_14): ∀x(G →F (x))→(G→∀x F (x)) $
@ -72,31 +56,30 @@ $bold(L_15): ∀x(F (x) arrow G) arrow (exists x F (x)→G)$
$[L_1, L_2, #[MP]]$: $[L_1, L_2, #[MP]]$:
+ $((A→B)→(A→C))→(A→(B→C))$. Be careful when assuming hypotheses: + $((A→B)→(A→C))→(A→(B→C))$. Be careful when assuming hypotheses: assume
assume (A→B)→(A→C), A, B – in this order, no other possibilities! (A→B)→(A→C), A, B – in this order, no other possibilities!
+ $(A→B)→((B→C)→(A→C))$. It's another version of the *Law of Syllogism* + $(A→B)→((B→C)→(A→C))$. It's another version of the *Law of Syllogism* (by
(by Aristotle), or the transitivity property of implication. Aristotle), or the transitivity property of implication.
+ $(A→(B→C))→(B→(A→C))$. It's another version of the *Premise + $(A→(B→C))→(B→(A→C))$. It's another version of the *Premise Permutation Law*.
Permutation Law*. Explain the difference between this formula and Theorem Explain the difference between this formula and Theorem 1.4.3(a): A→(B→C)├
1.4.3(a): A→(B→C)├ B→(A→C). B→(A→C).
== Deduction theorems == Deduction theorems
=== Theorem 1.5.1 (Deduction Theorem 1 AKA DT1) === Theorem 1.5.1 (Deduction Theorem 1 AKA DT1)
If $T$ is a first order theory, and there is a proof of If $T$ is a first order theory, and there is a proof of
$[T, #[MP]]: A_1, A_2, dots, A_n, B├ C$, $[T, #[MP]]: A_1, A_2, dots, A_n, B├ C$, then there is a proof of
then there is a proof of
$[L_1, L_2, T, #[MP]]: A_1, A_2, dots, A_n├ B→C$. $[L_1, L_2, T, #[MP]]: A_1, A_2, dots, A_n├ B→C$.
=== Theorem 1.5.2 (Deduction Theorem 2 AKA DT2) === Theorem 1.5.2 (Deduction Theorem 2 AKA DT2)
If there is a proof $[T, #[MP], #[Gen]]: A_1, A_2, dots, A_n, B├ C$, If there is a proof $[T, #[MP], #[Gen]]: A_1, A_2, dots, A_n, B├ C$, where,
where, after B appears in the proof, Generalization is not applied to the after B appears in the proof, Generalization is not applied to the variables
variables that occur as free in $B$, then there is a proof of that occur as free in $B$, then there is a proof of
$[L_1, L_2, L_14, T, #[MP], #[Gen]]: A_1, A_2, dots, A_n├ B→C$. $[L_1, L_2, L_14, T, #[MP], #[Gen]]: A_1, A_2, dots, A_n├ B→C$.
== Conjunction == Conjunction
=== Theorem 2.2.1. === Theorem 2.2.1.
@ -105,16 +88,17 @@ $[L_1, L_2, L_14, T, #[MP], #[Gen]]: A_1, A_2, dots, A_n├ B→C$.
=== Theorem 2.2.2. === Theorem 2.2.2.
+ $[L_1, L_2, L_5, #[MP]]: (A→(B→C)) ↔ ((A→B)→(A→C))$ (extension of the axiom L_2). + $[L_1, L_2, L_5, #[MP]]: (A→(B→C)) ↔ ((A→B)→(A→C))$ (extension of the axiom
+ $[L_1-L_4, #[MP]]: (A→B)∧( B→C)→( A→C)$ (another form of the *Law of L_2).
Syllogism*, or *transitivity property of implication*). + $[L_1-L_4, #[MP]]: (A→B)∧( B→C)→( A→C)$ (another form of the *Law of Syllogism*,
or *transitivity property of implication*).
=== Theorem 2.2.3 (properties of the conjunction connective). === Theorem 2.2.3 (properties of the conjunction connective).
$[L_1- L_5, #[MP]]$: $[L_1- L_5, #[MP]]$:
+ $A∧B↔B∧A$ . Conjunction is commutative. + $A∧B↔B∧A$ . Conjunction is commutative.
+ $ A∧(B∧C)↔( A∧B)∧C$. Conjunction is associative. + $ A∧(B∧C)↔( A∧B)∧C$. Conjunction is associative.
+ $A∧A↔A$ . Conjunction is idempotent. + $A∧A↔A$ . Conjunction is idempotent.
=== Theorem 2.2.4 (properties of the equivalence connective). === Theorem 2.2.4 (properties of the equivalence connective).
@ -125,26 +109,24 @@ $[L_1- L_5, #[MP]]$:
+ $(A↔B)→(B↔A)$ (symmetry), + $(A↔B)→(B↔A)$ (symmetry),
+ $(A↔B)→((B↔C) →((A↔C))$ (transitivity). + $(A↔B)→((B↔C) →((A↔C))$ (transitivity).
== Disjunction == Disjunction
=== Theorem 2.3.1 === Theorem 2.3.1
+ (D-introduction)$[L_6, L_7, #[MP]]: A├ A∨B; B├ A∨B$; + (D-introduction)$[L_6, L_7, #[MP]]: A├ A∨B; B├ A∨B$;
+ (D-elimination) If there is a proof $[T, #[MP]]: A_1, A_2, ..., A_n, B├ D$, + (D-elimination) If there is a proof $[T, #[MP]]: A_1, A_2, ..., A_n, B├ D$, and
and a proof $[T, #[MP]]: A_1, A_2, ..., A_n, C├ D$, then there is a proof $[T, a proof $[T, #[MP]]: A_1, A_2, ..., A_n, C├ D$, then there is a proof $[T,
L_1, L_2, L_8, #[MP]]: A_1, A_2, dots, A_n, B∨C ├ D$. L_1, L_2, L_8, #[MP]]: A_1, A_2, dots, A_n, B∨C ├ D$.
=== Theorem 2.3.2 === Theorem 2.3.2
a) $[ L_5, L_6-L_8, #[MP]]: A∨B↔B∨A$ . Disjunction is commutative. a) $[ L_5, L_6-L_8, #[MP]]: A∨B↔B∨A$ . Disjunction is commutative. b) $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨A↔A$ .
b) $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨A↔A$ . Disjunction is idempotent. Disjunction is idempotent.
=== Theorem 2.3.3. === Theorem 2.3.3.
Disjunction is associative: $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨(B∨C)↔( Disjunction is associative: $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨(B∨C)↔(
A∨B)∨C$. A∨B)∨C$.
=== Theorem 2.3.4. === Theorem 2.3.4.
@ -169,18 +151,16 @@ conjunction:
(N-elimination) If there is a proof (N-elimination) If there is a proof
$[T, #[MP]]: A_1, A_2, ..., A_n, B├ C$, and a proof $[T, #[MP]]: A_1, A_2, ..., A_n, $[T, #[MP]]: A_1, A_2, ..., A_n, B├ C$, and a proof $[T, #[MP]]: A_1, A_2, ..., A_n,
B├ ¬C$, then there is a proof $[T, L_1, L_2, L_9, #[MP]]: A_1, A_2, ..., A_n├ ¬B$. B├ ¬C$, then there is a proof $[T, L_1, L_2, L_9, #[MP]]: A_1, A_2, ..., A_n├ ¬B$.
=== Theorem 2.4.2. === Theorem 2.4.2.
a) $[L_1, L_2, L_9, #[MP]]: A, ¬B├ ¬(A→B)$. What does it mean? a) $[L_1, L_2, L_9, #[MP]]: A, ¬B├ ¬(A→B)$. What does it mean? b) $[L_1-L_4, L_9, #[MP]]: A∧¬B→¬( A→B)$.
b) $[L_1-L_4, L_9, #[MP]]: A∧¬B→¬( A→B)$.
=== Theorem 2.4.3. === Theorem 2.4.3.
$[L_1, L_2, L_9, #[MP]]: (A→B)→(¬B→¬A)$. What does it mean? $[L_1, L_2, L_9, #[MP]]: (A→B)→(¬B→¬A)$. What does it mean? It's the so-called
It's the so-called *Contraposition Law*. *Contraposition Law*.
Note. The following rule form of Contraposition Law is called *Modus Tollens*: Note. The following rule form of Contraposition Law is called *Modus Tollens*:
$[L_1, L_2, L_9, #[MP]]: A→B, ¬B├ ¬A, or, frac(A→B \; ¬B, ¬A)$ $[L_1, L_2, L_9, #[MP]]: A→B, ¬B├ ¬A, or, frac(A→B \; ¬B, ¬A)$
@ -191,11 +171,10 @@ $[L_1, L_2, L_9, #[MP]]: A→¬¬A$.
=== Theorem 2.4.5. === Theorem 2.4.5.
a) $[L_1, L_2, L_9, #[MP]]: ¬¬¬A↔¬A$. a) $[L_1, L_2, L_9, #[MP]]: ¬¬¬A↔¬A$. b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$.
b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$. What does it mean? This is a “weak What does it mean? This is a “weak form” of the *Law of Excluded Middle* that
form” of the *Law of Excluded Middle* that can be proved can be proved constructively. The formula $¬¬( A∨¬A)$ can be proved in the
constructively. The formula $¬¬( A∨¬A)$ can be proved in the constructive logic, constructive logic, but $A∨¬A$ can't – as we will see in Section 2.8.
but $A∨¬A$ can't – as we will see in Section 2.8.
=== Theorem 2.4.9. === Theorem 2.4.9.
@ -208,21 +187,20 @@ b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$. What does it mean? This i
=== Theorem 2.5.1. === Theorem 2.5.1.
+ $[L_1, L_8, L_10, #[MP]]: ¬A∨B→( A→B)$. + $[L_1, L_8, L_10, #[MP]]: ¬A∨B→( A→B)$.
+ $[L_1, L_2, L_6, #[MP]]: A∨B→(¬A→B) ├¬A→(A→B)$ . It means that the “natural” + $[L_1, L_2, L_6, #[MP]]: A∨B→(¬A→B) ├¬A→(A→B)$ . It means that the “natural”
rule $A∨B ;¬ A ├ B$ implies $L_10$! rule $A∨B ;¬ A ├ B$ implies $L_10$!
=== Theorem 2.5.2.
=== Theorem 2.5.2.
$[L_1-L_10, #[MP]]$: $[L_1-L_10, #[MP]]$:
+ $(¬¬A→¬¬B)→¬¬(A→B)$. It's the converse of Theorem 2.4.7(b). Hence, $[L_1-L_10, + $(¬¬A→¬¬B)→¬¬(A→B)$. It's the converse of Theorem 2.4.7(b). Hence, $[L_1-L_10,
#[MP]]:├ ¬¬(A→B)↔(¬¬A→¬¬B)$. #[MP]]:├ ¬¬(A→B)↔(¬¬A→¬¬B)$.
+ $¬¬A→(¬A→A)$. It's the converse of Theorem 2.4.6(a). Hence, [L1-L10, + $¬¬A→(¬A→A)$. It's the converse of Theorem 2.4.6(a). Hence, [L1-L10,
#[MP]]: $¬¬A↔(¬A→A)$. #[MP]]: $¬¬A↔(¬A→A)$.
+ $A∨¬ A→(¬¬A→A)$ . + $A∨¬ A→(¬¬A→A)$ .
+ $¬¬(¬¬A→A)$. What does it mean? It’s a “weak” form of the Double Negations + $¬¬(¬¬A→A)$. What does it mean? It’s a “weak” form of the Double Negations Law –
Law – provable in constructive logic. provable in constructive logic.
== Negation -- classical logic == Negation -- classical logic
@ -236,30 +214,25 @@ A$.
$[L_8, L_11, #[MP]]: A→B, ¬A→B├ B$. Or, by Deduction Theorem 1, $[L_1, L_2, L_8, $[L_8, L_11, #[MP]]: A→B, ¬A→B├ B$. Or, by Deduction Theorem 1, $[L_1, L_2, L_8,
L_11, #[MP]]: (A→B)→((¬A→B)→B)$. L_11, #[MP]]: (A→B)→((¬A→B)→B)$.
=== Theorem 2.6.3. === Theorem 2.6.3.
$[L_1-L_11, #[MP]]: (¬B→¬A)→(A→B)$. Hence, $[L_1-L_11, #[MP]]: (A→B) ↔ (¬B→¬A)$. $[L_1-L_11, #[MP]]: (¬B→¬A)→(A→B)$. Hence, $[L_1-L_11, #[MP]]: (A→B) ↔ (¬B→¬A)$.
=== Theorem 2.6.3.
=== Theorem 2.6.3.
_(another one with the same number of because numbering error (it seems like it))_ _(another one with the same number of because numbering error (it seems like it))_
$[L_1-L_9, L_11, #[MP]]: ˫ ¬(A∧B)→¬A∨¬B$ . Hence, $[L_1-L_9, L_11, #[MP]]: ˫ $[L_1-L_9, L_11, #[MP]]: ˫ ¬(A∧B)→¬A∨¬B$ . Hence, $[L_1-L_9, L_11, #[MP]]: ˫
¬(A∧B)↔¬A∨¬B$ . ¬(A∧B)↔¬A∨¬B$ .
=== Theorem 2.6.4. === Theorem 2.6.4.
$[L_1-L_8, L_11, #[MP]]: (A→B)→¬ A∨B $. Hence, (I-elimination) $[L_1-L_11, #[MP]]: $[L_1-L_8, L_11, #[MP]]: (A→B)→¬ A∨B $. Hence, (I-elimination) $[L_1-L_11, #[MP]]:
(A→B)↔¬ A∨B$. (A→B)↔¬ A∨B$.
=== Theorem 2.6.5. === Theorem 2.6.5.
$[L_1-L_11, #[MP]]: ¬(A→B)→A∧¬B $. $[L_1-L_11, #[MP]]: ¬(A→B)→A∧¬B $.
=== Theorem 2.7.1 (Glivenko's Theorem). === Theorem 2.7.1 (Glivenko's Theorem).
@ -276,7 +249,6 @@ L_11, #[MP]]$.
The axiom $L_9$ cannot be proved in $[L_1-L_8, L_10, #[MP]]$. The axiom $L_9$ cannot be proved in $[L_1-L_8, L_10, #[MP]]$.
== Replacement Theorem 1 == Replacement Theorem 1
Let us consider three formulas: $B$, $B'$, $C$, where $B$ is a sub-formula of Let us consider three formulas: $B$, $B'$, $C$, where $B$ is a sub-formula of
@ -290,39 +262,36 @@ $[L_1-L_9, #[MP]]: B↔B'├ C↔C'$.
Let us consider three formulas: $B$, $B'$, $C$, where $B$ is a sub-formula of Let us consider three formulas: $B$, $B'$, $C$, where $B$ is a sub-formula of
$C$, and $o(B)$ is any occurrence of $B$ in $C$. Let us denote by $C'$ the $C$, and $o(B)$ is any occurrence of $B$ in $C$. Let us denote by $C'$ the
formula obtained from $C$ by replacing $o(B)$ by B'. Then, in the minimal formula obtained from $C$ by replacing $o(B)$ by B'. Then, in the minimal logic,
logic,
$[L_1-L_9, L_12-L_15, #[MP], #[Gen]]: B↔B'├ C↔C'$. $[L_1-L_9, L_12-L_15, #[MP], #[Gen]]: B↔B'├ C↔C'$.
== Theorem 3.1.1. == Theorem 3.1.1.
$[L_1, L_2, L_12, L_13, #[MP]]: forall x B(x) arrow exists x B(x)$ . What does it $[L_1, L_2, L_12, L_13, #[MP]]: forall x B(x) arrow exists x B(x)$ . What does
mean? It prohibits "empty domains". it mean? It prohibits "empty domains".
Theorem 3.1.2. Theorem 3.1.2.
+ [L1, L2, L12, L14, MP, Gen]: ∀x(B→C)→(∀x B→∀xC). + [L1, L2, L12, L14, MP, Gen]: ∀x(B→C)→(∀x B→∀xC).
+ [L1, L2, L12-L15, MP, Gen]: ∀x(B→C)→(∃z(x+z+1=y).x B→∃z(x+z+1=y).xC). + [L1, L2, L12-L15, MP, Gen]: ∀x(B→C)→(∃z(x+z+1=y).x B→∃z(x+z+1=y).xC).
== Theorems 3.1.3. == Theorems 3.1.3.
If F is any formula, then: If F is any formula, then:
+ (U-introduction) [Gen]: F(x) ├∀x F(x) . + (U-introduction) [Gen]: F(x) ├∀x F(x) .
+ (U-elimination) [L12, MP, Gen]: ∀x F(x) ├F(x) . What does it mean? + (U-elimination) [L12, MP, Gen]: ∀x F(x) ├F(x) . What does it mean?
+ (E-introduction) [L13, MP, Gen]: F(x) ├∃z(x+z+1=y).x F(x) . What does it mean? + (E-introduction) [L13, MP, Gen]: F(x) ├∃z(x+z+1=y).x F(x) . What does it mean?
== Theorems 3.1.4.
== Theorems 3.1.4. If F is any formula, and G is a formula that does not contain free occurrences
of x, then:
If F is any formula, and G is a formula that does not contain free occurrences of x, then:
+ (U2-introduction) [L14, MP, Gen] G →F (x) ├G →∀x F (x) . What does it mean? + (U2-introduction) [L14, MP, Gen] G →F (x) ├G →∀x F (x) . What does it mean?
+ (E2-introduction) [L15, MP, Gen]: F (x)→G ├∃z(x+z+1=y).x F (x)→G . What does it mean? + (E2-introduction) [L15, MP, Gen]: F (x)→G ├∃z(x+z+1=y).x F (x)→G . What does it
mean?
== Theorem 3.1.5. == Theorem 3.1.5.
@ -330,37 +299,34 @@ If F is any formula, and G is a formula that does not contain free occurrences o
+ [L1, L2, L5, L13, L15, MP, Gen]: $exists x exists y B(x,y) ↔ exists y exists x B(x,y)$. + [L1, L2, L5, L13, L15, MP, Gen]: $exists x exists y B(x,y) ↔ exists y exists x B(x,y)$.
+ [L1, L2, L12-L15, MP, Gen]: $exists x forall y B(x,y) ↔ forall y exists x B(x,y)$. + [L1, L2, L12-L15, MP, Gen]: $exists x forall y B(x,y) ↔ forall y exists x B(x,y)$.
Theorem 3.1.6. If the formula B does not contain free occurrences of x, then Theorem 3.1.6. If the formula B does not contain free occurrences of x, then
[L1-L2, L12-L15, MP, Gen]: (∀x B)↔B;(∃z(x+z+1=y).x B)↔B , i.e., quantifiers [L1-L2, L12-L15, MP, Gen]: (∀x B)↔B;(∃z(x+z+1=y).x B)↔B , i.e., quantifiers ∀x
∀x ;∃z(x+z+1=y). x can be dropped or introduced as needed. ;∃z(x+z+1=y). x can be dropped or introduced as needed.
Theorem 3.2.1. In the classical logic, Theorem 3.2.1. In the classical logic, [L1-L15, MP, Gen]: ¬ x¬B ∀ ↔ xB.
[L1-L15, MP, Gen]: ¬
x¬B
∀
↔ xB.
== Theorem 3.3.1.
== Theorem 3.3.1.
+ [L1-L5, L12, L14, MP, Gen]: ∀x(B∧C)↔∀x B∧∀xC . + [L1-L5, L12, L14, MP, Gen]: ∀x(B∧C)↔∀x B∧∀xC .
+ [L1, L2, L6-L8, L12, L14, MP, Gen]: ├∀x B∨∀xC →∀x(B∨C) . The converse formula ∀x(B∨C)→∀x B∨∀xC cannot be true. Explain, why. + [L1, L2, L6-L8, L12, L14, MP, Gen]: ├∀x B∨∀xC →∀x(B∨C) . The converse formula
∀x(B∨C)→∀x B∨∀xC cannot be true. Explain, why.
== Theorem 3.3.2. == Theorem 3.3.2.
a) [L1-L8, L12-L15, MP, Gen]: ∃z(x+z+1=y).x(B∨C)↔∃z(x+z+1=y). x B∨∃z(x+z+1=y).xC . a) [L1-L8, L12-L15, MP, Gen]: ∃z(x+z+1=y).x(B∨C)↔∃z(x+z+1=y). x B∨∃z(x+z+1=y).xC
b) [L1-L5, L13-L15, MP, Gen]: ∃z(x+z+1=y).x(B∧C)→∃z(x+z+1=y). x B∧∃z(x+z+1=y).xC . The converse implication ∃z(x+z+1=y).x B∧∃z(x+z+1=y). xC →∃z(x+z+1=y). x(B∧C) cannot be true. Explain, why. Exercise 3.3.3. a) Prove (a→) of Theorem 3.3.2. (Hint: start by assuming B∨C , apply D-elimination, etc., and finish by E2-introduction.) . b) [L1-L5, L13-L15, MP, Gen]: ∃z(x+z+1=y).x(B∧C)→∃z(x+z+1=y). x
B∧∃z(x+z+1=y).xC . The converse implication ∃z(x+z+1=y).x B∧∃z(x+z+1=y). xC
→∃z(x+z+1=y). x(B∧C) cannot be true. Explain, why. Exercise 3.3.3. a) Prove (a→)
of Theorem 3.3.2. (Hint: start by assuming B∨C , apply D-elimination, etc., and
finish by E2-introduction.)
= Three-valued logic = Three-valued logic
This is a general scheme (page 74) to define a three valued logic. This is a general scheme (page 74) to define a three valued logic.
For example, let us consider a kind of "three-valued logic", where 0 means For example, let us consider a kind of "three-valued logic", where 0 means
"false", 1 – "unknown" (or NULL – in terms of SQL), and 2 means "true". "false", 1 – "unknown" (or NULL – in terms of SQL), and 2 means "true". Then it
Then it would be natural to define “truth values” of conjunction and would be natural to define “truth values” of conjunction and disjunction as
disjunction as
$A∧B=min ( A, B)$ ; $A∧B=min ( A, B)$ ;
@ -369,25 +335,11 @@ $A∨B=max (A , B)$ .
But how should we define “truth values” of implication and negation? But how should we define “truth values” of implication and negation?
#table( #table(
columns: 5, columns: 5, [$A$], [$B$], [$A∧B$], [$A∨B$], [$A→B$], [$0$], [$0$], [$0$], [$0$], [$i_1$], [$0$], [$1$], [$0$], [$1$], [$i_2$], [$0$], [$2$], [$0$], [$2$], [$i_3$], [$1$], [$0$], [$0$], [$1$], [$i_4$], [$1$], [$1$], [$1$], [$1$], [$i_5$], [$1$], [$2$], [$1$], [$2$], [$i_6$], [$2$], [$0$], [$0$], [$2$], [$i_7$], [$2$], [$1$], [$1$], [$2$], [$i_8$], [$2$], [$2$], [$2$], [$2$], [$i_9$],
[$A$], [$B$], [$A∧B$], [$A∨B$], [$A→B$],
[$0$], [$0$], [$0$], [$0$], [$i_1$],
[$0$], [$1$], [$0$], [$1$], [$i_2$],
[$0$], [$2$], [$0$], [$2$], [$i_3$],
[$1$], [$0$], [$0$], [$1$], [$i_4$],
[$1$], [$1$], [$1$], [$1$], [$i_5$],
[$1$], [$2$], [$1$], [$2$], [$i_6$],
[$2$], [$0$], [$0$], [$2$], [$i_7$],
[$2$], [$1$], [$1$], [$2$], [$i_8$],
[$2$], [$2$], [$2$], [$2$], [$i_9$],
) )
#table( #table(
columns: 2, columns: 2, [$A$], [¬$A$], [$0$], [$i_10$], [$1$], [$i_11$], [$2$], [$i_12$],
[$A$], [¬$A$],
[$0$], [$i_10$],
[$1$], [$i_11$],
[$2$], [$i_12$],
) )
= Model interpreation = Model interpreation
@ -400,28 +352,27 @@ Let L be a predicate language containing:
- (a possibly empty) set of object constants $c_1, dots, c_k, dots $; - (a possibly empty) set of object constants $c_1, dots, c_k, dots $;
- (a possibly empty) set of function constants $f_1, dots, f_m, dots,$; - (a possibly empty) set of function constants $f_1, dots, f_m, dots,$;
- (a non empty) set of predicate constants $p_1, ..., p_n, ...$. - (a non empty) set of predicate constants $p_1, ..., p_n, ...$.
An interpretation $J$ of the language $L$ consists of the following two An interpretation $J$ of the language $L$ consists of the following two entities
entities (a set and a mapping): (a set and a mapping):
+ A non-empty finite or infinite set DJ – the domain of interpretation (it will + A non-empty finite or infinite set DJ – the domain of interpretation (it will
serve first of all as the range of object variables). (For infinite domains, set serve first of all as the range of object variables). (For infinite domains, set
theory comes in here.) theory comes in here.)
+ A mapping intJ that assigns: + A mapping intJ that assigns:
- to each object constant $c_i$ – a member $#[int]_J (c_i)$ of the domain - to each object constant $c_i$ – a member $#[int]_J (c_i)$ of the domain
$D_J$ [contstant corresponds to an object from domain]; $D_J$ [contstant corresponds to an object from domain];
- to each function constant $f_i$ – a function $#[int]_J (f_i)$ from $D_J times dots - to each function constant $f_i$ – a function $#[int]_J (f_i)$ from $D_J times dots
times D_J$ into $D_J$ [], times D_J$ into $D_J$ [],
- to each predicate constant $p_i$ – a predicate $#[int]_J (p_i)$ on $D_J$. - to each predicate constant $p_i$ – a predicate $#[int]_J (p_i)$ on $D_J$.
Having an interpretation $J$ of the language $L$, we can define the notion of Having an interpretation $J$ of the language $L$, we can define the notion of
*true formulas* (more precisely − the notion of formulas that are true under *true formulas* (more precisely − the notion of formulas that are true under the
the interpretation $J$). interpretation $J$).
*Example.* The above interpretation of the “language about people” put in the *Example.* The above interpretation of the “language about people” put in the
terms of the general definition: terms of the general definition:
+ $D = {#[br], #[jo], #[pa], #[pe]}$. + $D = {#[br], #[jo], #[pa], #[pe]}$.
+ $#[int]_J (#[Britney])=#[br], #[int]_J (#[John])=#[jo], #[int]_J (#[Paris])=#[pa], + $#[int]_J (#[Britney])=#[br], #[int]_J (#[John])=#[jo], #[int]_J (#[Paris])=#[pa],
@ -433,40 +384,38 @@ terms of the general definition:
+ $#[int]_J (=) = {(#[br], #[br]), (#[jo], #[jo]), (#[pa], #[pa]), (#[pe], + $#[int]_J (=) = {(#[br], #[br]), (#[jo], #[jo]), (#[pa], #[pa]), (#[pe],
#[pe])}$. #[pe])}$.
=== Interpretations of languages − the standard common part === Interpretations of languages − the standard common part
Finally, we define the notion of *true formulas* of the language $L$ under the Finally, we define the notion of *true formulas* of the language $L$ under the
interpretation $J$ (of course, for a fixed combination of values of their free interpretation $J$ (of course, for a fixed combination of values of their free
variables – if any): variables – if any):
+ Truth-values of the formulas: $¬B , B∧C , B∨C , B →C$ [those are not + Truth-values of the formulas: $¬B , B∧C , B∨C , B →C$ [those are not examples]
examples] must be computed. This is done with the truth-values of $B$ and $C$ must be computed. This is done with the truth-values of $B$ and $C$
by using the well-known classical truth tables (see Section 4.2). by using the well-known classical truth tables (see Section 4.2).
+ The formula $∀x B$ is true under $J$ if and only if $B(c)$ is true under $J$ + The formula $∀x B$ is true under $J$ if and only if $B(c)$ is true under $J$
for all members $c$ of the domain $D_J$. for all members $c$ of the domain $D_J$.
+ The formula $∃x B$ is true under $J$ if and only if there is a member c of the + The formula $∃x B$ is true under $J$ if and only if there is a member c of the
domain $D_J$ such that $B(c)$ is true under $J$. domain $D_J$ such that $B(c)$ is true under $J$.
*Example.* In first order arithmetic, the formula *Example.* In first order arithmetic, the formula
$ $
y((x= y+ y)∨( x=y+ y+1)) y((x= y+ y)∨( x=y+ y+1))
$ $
is intended to say that "x is even or odd". Under the standard interpretation S is intended to say that "x is even or odd". Under the standard interpretation S
of arithmetic, this formula is true for all values of its free variable x. of arithmetic, this formula is true for all values of its free variable x.
Similarly, $∀x ∀y(x+ y=y+x)$ is a closed formula that is true under Similarly, $∀x ∀y(x+ y=y+x)$ is a closed formula that is true under this
this interpretation. The notion “a closed formula F is true under the interpretation. The notion “a closed formula F is true under the interpretation
interpretation J” is now precisely defined. J” is now precisely defined.
*Important − non-constructivity!* It may seem that, under an interpretation, any
*Important − non-constructivity!* It may seem that, under an interpretation, closed formula is "either true or false". However, note that, for an infinite
any closed formula is "either true or false". However, note that, for an infinite domain DJ, the notion of "true formulas under J" is extremely non- constructive.
domain DJ, the notion of "true formulas under J" is extremely non-
constructive.
=== Example of building of an interpretation === Example of building of an interpretation
@ -474,7 +423,7 @@ In Section 1.2, in our "language about people" we used four names of people
(Britney, John, Paris, Peter) as object constants and the following predicate (Britney, John, Paris, Peter) as object constants and the following predicate
constants: constants:
+ $#[Male] (x)$ − means "x is a male person"; + $#[Male] (x)$ − means "x is a male person";
+ $#[Female] (x)$ − means "x is a female person"; + $#[Female] (x)$ − means "x is a female person";
+ $#[Mother] (x, y)$ − means "x is mother of y"; + $#[Mother] (x, y)$ − means "x is mother of y";
+ $#[Father] (x, y)$ − means "x is father of y"; + $#[Father] (x, y)$ − means "x is father of y";
@ -491,55 +440,32 @@ Interpretations of predicate constants are defined by the following truth
tables: tables:
#table( #table(
columns: 3, columns: 3, [x], [Male(x)], [Female(x)], [br], [false], [true], [jo], [true], [false], [pa], [false], [true], [pe], [true], [false],
[x], [Male(x)], [Female(x)],
[br], [false], [true],
[jo], [true], [false],
[pa], [false], [true],
[pe], [true], [false],
) )
#table( #table(
columns: 6, columns: 6, [x], [y], [Father(x,y)], [Mother(x,y)], [Married(x,y)], [x=y], [br], [br], [false], [false], [false], [true], [br], [jo], [false], [false], [false], [false], [br], [pa], [false], [false], [false], [false], [br], [pe], [false], [false], [false], [false], [jo], [br], [false], [false], [false], [false], [jo], [jo], [false], [false], [false], [true], [jo], [pa], [false], [false], [false], [false], [jo], [pe], [false], [false], [false], [false], [pa], [br], [false], [true], [false], [false], [pa], [jo], [false], [true], [false], [false], [pa], [pa], [false], [false], [false], [true], [pa], [pe], [false], [false], [true], [false], [pe], [br], [true], [false], [false], [false], [pe], [jo], [true], [false], [false], [false], [pe], [pa], [false], [false], [true], [false], [pe], [pe], [false], [false], [false], [true],
[x], [y], [Father(x,y)], [Mother(x,y)], [Married(x,y)], [x=y],
[br], [br], [false], [false], [false], [true],
[br], [jo], [false], [false], [false], [false],
[br], [pa], [false], [false], [false], [false],
[br], [pe], [false], [false], [false], [false],
[jo], [br], [false], [false], [false], [false],
[jo], [jo], [false], [false], [false], [true],
[jo], [pa], [false], [false], [false], [false],
[jo], [pe], [false], [false], [false], [false],
[pa], [br], [false], [true], [false], [false],
[pa], [jo], [false], [true], [false], [false],
[pa], [pa], [false], [false], [false], [true],
[pa], [pe], [false], [false], [true], [false],
[pe], [br], [true], [false], [false], [false],
[pe], [jo], [true], [false], [false], [false],
[pe], [pa], [false], [false], [true], [false],
[pe], [pe], [false], [false], [false], [true],
) )
== Three kinds of formulas == Three kinds of formulas
If one explores some formula F of the language L under various If one explores some formula F of the language L under various interpretations,
interpretations, then three situations are possible: then three situations are possible:
+ $F$ is true in all interpretations of the language $L$. Formulas of this kind are + $F$ is true in all interpretations of the language $L$. Formulas of this kind
called *logically valid formulas* (LVF, Latv. *LVD*). are called *logically valid formulas* (LVF, Latv. *LVD*).
+ $F$ is true in some interpretations of $L$, and false − in some other + $F$ is true in some interpretations of $L$, and false − in some other
interpretations of $L$. interpretations of $L$.
+ F is false in all interpretations of L Formulas of this kind are called + F is false in all interpretations of L Formulas of this kind are called
*unsatisfiable formulas* (Latv. *neizpildāmas funkcijas*). *unsatisfiable formulas* (Latv. *neizpildāmas funkcijas*).
Formulas that are "not unsatisfiable" (formulas of classes (a) and (b)) are Formulas that are "not unsatisfiable" (formulas of classes (a) and (b)) are
called, of course, satisfiable formulas: a formula is satisfiable, if it is called, of course, satisfiable formulas: a formula is satisfiable, if it is true
true in at least one interpretation [*satisfiable functions* (Latv. *izpildāmas in at least one interpretation [*satisfiable functions* (Latv. *izpildāmas
funkcijas*)]. funkcijas*)].
=== Prooving an F is LVF (Latv. LVD) === Prooving an F is LVF (Latv. LVD)
First, we should learn to prove that some formula is (if really is!) logically First, we should learn to prove that some formula is (if really is!) logically
@ -551,46 +477,41 @@ logically valid. Check pages 125-126 of the book for example of such proof
check it, because in such way you will need to solve tasks in homeworks and check it, because in such way you will need to solve tasks in homeworks and
tests. tests.
=== Prooving an F is satisfiable but NOT LVF === Prooving an F is satisfiable but NOT LVF
As an example, let us verify that the formula As an example, let us verify that the formula
$ $
∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x) ∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)
$ $
is not logically valid (p, q are predicate constants). Why it is not? Because is not logically valid (p, q are predicate constants). Why it is not? Because
the truth-values of p(x) and q(x) may behave in such a way that $p(x)∨q(x)$ is the truth-values of p(x) and q(x) may behave in such a way that $p(x)∨q(x)$ is
always true, but neither $forall x p(x)$, nor $forall x q(x)$ is true. Indeed, always true, but neither $forall x p(x)$, nor $forall x q(x)$ is true. Indeed,
let us take the domain $D = {a, b}$, and set (in fact, we are using one of two let us take the domain $D = {a, b}$, and set (in fact, we are using one of two
possibilities): possibilities):
#table( #table(
columns: 3, columns: 3, [x], [p(x)], [q(x)], [b], [false], [true], [a], [true], [false],
[x], [p(x)], [q(x)],
[b], [false], [true],
[a], [true], [false],
) )
In this interpretation, $p(a)∨q(a) = #[true]$ , $p(b)∨q(b) = #[true]$, In this interpretation, $p(a)∨q(a) = #[true]$ , $p(b)∨q(b) = #[true]$, i.e., the
i.e., the premise $∀x( p( x)∨q(x))$ is true. But the formulas$forall p(x), premise $∀x( p( x)∨q(x))$ is true. But the formulas$forall p(x),
forall q(x)$ both are false. Hence, in this interpretation, the conclusion $∀x forall q(x)$ both are false. Hence, in this interpretation, the conclusion $∀x
p(x)∨∀x q(x)$ is false, and $∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)$ is false. We p(x)∨∀x q(x)$ is false, and $∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)$ is false. We have
have built an interpretation, making the formula false. Hence, it is not built an interpretation, making the formula false. Hence, it is not logically
logically valid. valid.
On the other hand, this formula is satisfiable – there is an interpretation On the other hand, this formula is satisfiable – there is an interpretation
under which it is true. Indeed, let us take $D={a}$ as the domain of under which it is true. Indeed, let us take $D={a}$ as the domain of
interpretation, and let us set $p(a)=q(a)=#[true]$. Then all the formulas interpretation, and let us set $p(a)=q(a)=#[true]$. Then all the formulas
$ $
∀x( p( x)∨q( x)),∀x p(x),∀x q( x) ∀x( p( x)∨q( x)),∀x p(x),∀x q( x)
$ $
become true, and so becomes the entire formula. become true, and so becomes the entire formula.
== Gödel's Completeness Theorem == Gödel's Completeness Theorem
*Theorem 4.3.1.* In classical predicate logic $[L_1−L_15,#[MP],#[Gen]]$ all *Theorem 4.3.1.* In classical predicate logic $[L_1−L_15,#[MP],#[Gen]]$ all
@ -610,7 +531,6 @@ can make it false). If we manage to do so, then we can announce: according to
Gödel’s theorem, $F$ is derivable in classical predicate logic Gödel’s theorem, $F$ is derivable in classical predicate logic
$[L_1−L_15,#[MP],#[Gen]]$. $[L_1−L_15,#[MP],#[Gen]]$.
= Tableaux algorithm = Tableaux algorithm
== Step 1. == Step 1.
@ -618,9 +538,9 @@ $[L_1−L_15,#[MP],#[Gen]]$.
We will solve the task from the opposite: append to the hypotheses $F_1, dots We will solve the task from the opposite: append to the hypotheses $F_1, dots
F_n$ negation of formula $G$, and obtain the set $F_1, dots, F_n, ¬G$. When you F_n$ negation of formula $G$, and obtain the set $F_1, dots, F_n, ¬G$. When you
will do homework or test, you shouldn’t forget this, because if you work with will do homework or test, you shouldn’t forget this, because if you work with
the set $F_1, ..., F_n, G$, then obtained result will not give an answer the set $F_1, ..., F_n, G$, then obtained result will not give an answer whether $G$ is
whether $G$ is derivable or not. You should keep this in mind also when the derivable or not. You should keep this in mind also when the task has only one
task has only one formula, e.g., verify, whether formula $(A→B)→((B→C)→(A→C))$ formula, e.g., verify, whether formula $(A→B)→((B→C)→(A→C))$
is derivable. Then from the beginning you should append negation in front: is derivable. Then from the beginning you should append negation in front:
¬((A→B)→((B→C)→(A→C))) and then work further. Instead of the set $F_1, dots, ¬((A→B)→((B→C)→(A→C))) and then work further. Instead of the set $F_1, dots,
F_n, ¬G$ we can always check one formula $F_1∧...∧F_n∧¬G$. Therefore, our task F_n, ¬G$ we can always check one formula $F_1∧...∧F_n∧¬G$. Therefore, our task
@ -635,21 +555,21 @@ Substitution theorem. First, implications are replaced with negations and
disjunctions: disjunctions:
$ $
(A→B)↔¬A∨B (A→B)↔¬A∨B
$ $
Then we apply de Morgan laws to get negations close to the atoms: Then we apply de Morgan laws to get negations close to the atoms:
$ $
¬(A∨B)↔¬A∧¬B equiv \ ¬(A∨B)↔¬A∧¬B equiv \
¬(A∧B)↔¬A∨¬B ¬(A∧B)↔¬A∨¬B
$ $
In such way all negations are carried exactly before atoms. In such way all negations are carried exactly before atoms. After that we can
After that we can remove double negations: remove double negations:
$ $
¬¬A↔A ¬¬A↔A
$ $
Example: $(p→q)→q$. Example: $(p→q)→q$.
@ -666,24 +586,24 @@ atoms.
== Step 3. == Step 3.
Next, we should build a tree, vertices of which are formulas. In the root of Next, we should build a tree, vertices of which are formulas. In the root of the
the tree we put our formula. Then we have two cases. tree we put our formula. Then we have two cases.
+ If vertex is formula A∧B, then each branch that goes through this vertex is + If vertex is formula A∧B, then each branch that goes through this vertex is
extended with vertices A and B. extended with vertices A and B.
+ If vertex is a formula A∨B, then in place of continuation we have branching + If vertex is a formula A∨B, then in place of continuation we have branching into
into vertex A and vertex B. vertex A and vertex B.
In both cases, the initial vertex is marked as processed. Algorithm continues In both cases, the initial vertex is marked as processed. Algorithm continues to
to process all cases 1 and 2 until all non-atomic vertices have been processed. process all cases 1 and 2 until all non-atomic vertices have been processed.
== Step 4. == Step 4.
When the construction of the tree is finished, we need to analyze and make When the construction of the tree is finished, we need to analyze and make
conclusions. When one branch has some atom both with and without a negation conclusions. When one branch has some atom both with and without a negation
(e.g., $A$ and $¬A$), then it is called closed branch. Other branches are (e.g., $A$ and $¬A$), then it is called closed branch. Other branches are called
called open branches. open branches.
*Theorem.* If in constructed tree, there exists at least one open branch, then *Theorem.* If in constructed tree, there exists at least one open branch, then
formula in the root is satisfiable. And vice versa – if all branches in the formula in the root is satisfiable. And vice versa – if all branches in the tree
tree are closed, then formula in the root is unsatisfiable. are closed, then formula in the root is unsatisfiable.