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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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LICENSE Normal file
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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(
margin: (
left: 1cm,
right: 1cm,
top: 1cm,
bottom: 1cm,
),
)
#set page(margin: 1cm, columns: 2)
#set heading(numbering: "1.")
#show outline.entry.where(
level: 1
): it => {
#show outline.entry.where(level: 1): it => {
v(12pt, weak: true)
strong(it)
}
#set enum(numbering: "a)")
// #show outline.entry.where(
@ -31,14 +18,11 @@
#outline(indent: auto)
= Logical axiom schemes
$bold(L_1): B→(C →B)$
$bold(L_2): (B→(C →D))→((B→C)→( B→D))$
$bold(L_2): (B→(C →D))→((B→C)→( B→D))$
$bold(L_3): B∧C→B$
@ -58,9 +42,9 @@ $bold(L_10): ¬B→( B→C)$
$bold(L_11): B∨¬B$
$bold(L_12): ∀x F (x)→F (t)$ (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)) $
@ -72,28 +56,27 @@ $bold(L_15): ∀x(F (x) arrow G) arrow (exists x F (x)→G)$
$[L_1, L_2, #[MP]]$:
+ $((A→B)→(A→C))→(A→(B→C))$. Be careful when assuming hypotheses:
assume (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*
(by Aristotle), or the transitivity property of implication.
+ $(A→(B→C))→(B→(A→C))$. It's another version of the *Premise
Permutation Law*. Explain the difference between this formula and Theorem
1.4.3(a): A→(B→C)├ B→(A→C).
+ $((A→B)→(A→C))→(A→(B→C))$. Be careful when assuming hypotheses: assume
(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* (by
Aristotle), or the transitivity property of implication.
+ $(A→(B→C))→(B→(A→C))$. It's another version of the *Premise Permutation Law*.
Explain the difference between this formula and Theorem 1.4.3(a): A→(B→C)├
B→(A→C).
== Deduction theorems
=== Theorem 1.5.1 (Deduction Theorem 1 AKA DT1)
If $T$ is a first order theory, and there is a proof of
$[T, #[MP]]: A_1, A_2, dots, A_n, B├ C$,
then there is a proof of
$[T, #[MP]]: A_1, A_2, dots, A_n, B├ C$, then there is a proof of
$[L_1, L_2, T, #[MP]]: A_1, A_2, dots, A_n├ B→C$.
=== Theorem 1.5.2 (Deduction Theorem 2 AKA DT2)
If there is a proof $[T, #[MP], #[Gen]]: A_1, A_2, dots, A_n, B├ C$,
where, after B appears in the proof, Generalization is not applied to the
variables that occur as free in $B$, then there is a proof of
If there is a proof $[T, #[MP], #[Gen]]: A_1, A_2, dots, A_n, B├ C$, where,
after B appears in the proof, Generalization is not applied to the variables
that occur as free in $B$, then there is a proof of
$[L_1, L_2, L_14, T, #[MP], #[Gen]]: A_1, A_2, dots, A_n├ B→C$.
== Conjunction
@ -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.
+ $[L_1, L_2, L_5, #[MP]]: (A→(B→C)) ↔ ((A→B)→(A→C))$ (extension of the axiom L_2).
+ $[L_1-L_4, #[MP]]: (A→B)∧( B→C)→( A→C)$ (another form of the *Law of
Syllogism*, or *transitivity property of implication*).
+ $[L_1, L_2, L_5, #[MP]]: (A→(B→C)) ↔ ((A→B)→(A→C))$ (extension of the axiom
L_2).
+ $[L_1-L_4, #[MP]]: (A→B)∧( B→C)→( A→C)$ (another form of the *Law of Syllogism*,
or *transitivity property of implication*).
=== Theorem 2.2.3 (properties of the conjunction connective).
$[L_1- L_5, #[MP]]$:
+ $A∧B↔B∧A$ . 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.
=== Theorem 2.2.4 (properties of the equivalence connective).
@ -125,26 +109,24 @@ $[L_1- L_5, #[MP]]$:
+ $(A↔B)→(B↔A)$ (symmetry),
+ $(A↔B)→((B↔C) →((A↔C))$ (transitivity).
== Disjunction
=== Theorem 2.3.1
+ (D-introduction)$[L_6, L_7, #[MP]]: A├ A∨B; B├ A∨B$;
+ (D-elimination) If there is a proof $[T, #[MP]]: A_1, A_2, ..., A_n, B├ D$,
and a proof $[T, #[MP]]: A_1, A_2, ..., A_n, C├ D$, then there is a proof $[T,
L_1, L_2, L_8, #[MP]]: A_1, A_2, dots, A_n, B∨C ├ D$.
+ (D-elimination) If there is a proof $[T, #[MP]]: A_1, A_2, ..., A_n, B├ D$, and
a proof $[T, #[MP]]: A_1, A_2, ..., A_n, C├ D$, then there is a proof $[T,
L_1, L_2, L_8, #[MP]]: A_1, A_2, dots, A_n, B∨C ├ D$.
=== Theorem 2.3.2
a) $[ L_5, L_6-L_8, #[MP]]: A∨B↔B∨A$ . Disjunction is commutative.
b) $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨A↔A$ . Disjunction is idempotent.
a) $[ L_5, L_6-L_8, #[MP]]: A∨B↔B∨A$ . Disjunction is commutative. b) $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨A↔A$ .
Disjunction is idempotent.
=== Theorem 2.3.3.
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.
@ -169,18 +151,16 @@ conjunction:
(N-elimination) If there is a proof
$[T, #[MP]]: A_1, A_2, ..., A_n, B├ C$, and a proof $[T, #[MP]]: A_1, A_2, ..., A_n,
B├ ¬C$, 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.
a) $[L_1, L_2, L_9, #[MP]]: A, ¬B├ ¬(A→B)$. What does it mean?
b) $[L_1-L_4, L_9, #[MP]]: A∧¬B→¬( A→B)$.
a) $[L_1, L_2, L_9, #[MP]]: A, ¬B├ ¬(A→B)$. What does it mean? b) $[L_1-L_4, L_9, #[MP]]: A∧¬B→¬( A→B)$.
=== Theorem 2.4.3.
$[L_1, L_2, L_9, #[MP]]: (A→B)→(¬B→¬A)$. What does it mean?
It's the so-called *Contraposition Law*.
$[L_1, L_2, L_9, #[MP]]: (A→B)→(¬B→¬A)$. What does it mean? It's the so-called
*Contraposition Law*.
Note. The following rule form of Contraposition Law is called *Modus Tollens*:
$[L_1, L_2, L_9, #[MP]]: A→B, ¬B├ ¬A, or, frac(A→B \; ¬B, ¬A)$
@ -191,11 +171,10 @@ $[L_1, L_2, L_9, #[MP]]: A→¬¬A$.
=== Theorem 2.4.5.
a) $[L_1, L_2, L_9, #[MP]]: ¬¬¬A↔¬A$.
b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$. What does it mean? This is a “weak
form” of the *Law of Excluded Middle* that can be proved
constructively. The formula $¬¬( A∨¬A)$ can be proved in the constructive logic,
but $A∨¬A$ can't – as we will see in Section 2.8.
a) $[L_1, L_2, L_9, #[MP]]: ¬¬¬A↔¬A$. b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$.
What does it mean? This is a “weak form” of the *Law of Excluded Middle* that
can be proved constructively. The formula $¬¬( A∨¬A)$ can be proved in the
constructive logic, but $A∨¬A$ can't – as we will see in Section 2.8.
=== 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.
+ $[L_1, L_8, L_10, #[MP]]: ¬A∨B→( A→B)$.
+ $[L_1, L_2, L_6, #[MP]]: A∨B→(¬A→B) ├¬A→(A→B)$ . 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 $L_10$!
=== Theorem 2.5.2.
$[L_1-L_10, #[MP]]$:
+ $(¬¬A→¬¬B)→¬¬(A→B)$. It's the converse of Theorem 2.4.7(b). Hence, $[L_1-L_10,
#[MP]]:├ ¬¬(A→B)↔(¬¬A→¬¬B)$.
+ $¬¬A→(¬A→A)$. 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]]: $¬¬A↔(¬A→A)$.
+ $A∨¬ A→(¬¬A→A)$ .
+ $¬¬(¬¬A→A)$. What does it mean? It’s a “weak” form of the Double Negations
Law – provable in constructive logic.
+ $¬¬(¬¬A→A)$. What does it mean? It’s a “weak” form of the Double Negations Law –
provable in constructive logic.
== Negation -- classical logic
@ -236,12 +214,10 @@ A$.
$[L_8, L_11, #[MP]]: A→B, ¬A→B├ B$. Or, by Deduction Theorem 1, $[L_1, L_2, L_8,
L_11, #[MP]]: (A→B)→((¬A→B)→B)$.
=== Theorem 2.6.3.
$[L_1-L_11, #[MP]]: (¬B→¬A)→(A→B)$. Hence, $[L_1-L_11, #[MP]]: (A→B) ↔ (¬B→¬A)$.
=== Theorem 2.6.3.
_(another one with the same number of because numbering error (it seems like it))_
@ -249,18 +225,15 @@ _(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, $[L_1-L_9, L_11, #[MP]]: ˫
¬(A∧B)↔¬A∨¬B$ .
=== Theorem 2.6.4.
$[L_1-L_8, L_11, #[MP]]: (A→B)→¬ A∨B $. Hence, (I-elimination) $[L_1-L_11, #[MP]]:
(A→B)↔¬ A∨B$.
=== Theorem 2.6.5.
$[L_1-L_11, #[MP]]: ¬(A→B)→A∧¬B $.
=== Theorem 2.7.1 (Glivenko's Theorem).
$[L_1-L_11, #[MP]]:├ A$ if and only if $[L_1-L_10, #[MP]]:├ ¬¬A$.
@ -276,7 +249,6 @@ L_11, #[MP]]$.
The axiom $L_9$ cannot be proved in $[L_1-L_8, L_10, #[MP]]$.
== Replacement Theorem 1
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
$C$, 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
logic,
formula obtained from $C$ by replacing $o(B)$ by B'. Then, in the minimal logic,
$[L_1-L_9, L_12-L_15, #[MP], #[Gen]]: B↔B'├ C↔C'$.
== 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
mean? It prohibits "empty domains".
$[L_1, L_2, L_12, L_13, #[MP]]: forall x B(x) arrow exists x B(x)$ . What does
it mean? It prohibits "empty domains".
Theorem 3.1.2.
+ [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).
== Theorems 3.1.3.
If F is any formula, then:
+ (U-introduction) [Gen]: F(x) ├∀x F(x) .
+ (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?
+ (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?
== 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?
+ (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.
@ -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, 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
[L1-L2, L12-L15, MP, Gen]: (∀x B)↔B;(∃z(x+z+1=y).x B)↔B , i.e., quantifiers
∀x ;∃z(x+z+1=y). x can be dropped or introduced as needed.
Theorem 3.2.1. In the classical logic,
[L1-L15, MP, Gen]: ¬
x¬B
∀
↔ xB.
[L1-L2, L12-L15, MP, Gen]: (∀x B)↔B;(∃z(x+z+1=y).x B)↔B , i.e., quantifiers ∀x
;∃z(x+z+1=y). x can be dropped or introduced as needed.
Theorem 3.2.1. In the classical logic, [L1-L15, MP, Gen]: ¬ x¬B ∀ ↔ xB.
== Theorem 3.3.1.
+ [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.
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.)
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.)
= 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
"false", 1 – "unknown" (or NULL – in terms of SQL), and 2 means "true".
Then it would be natural to define “truth values” of conjunction and
disjunction as
"false", 1 – "unknown" (or NULL – in terms of SQL), and 2 means "true". Then it
would be natural to define “truth values” of conjunction and disjunction as
$A∧B=min ( A, B)$ ;
@ -369,25 +335,11 @@ $A∨B=max (A , B)$ .
But how should we define “truth values” of implication and negation?
#table(
columns: 5,
[$A$], [$B$], [$A∧B$], [$A∨B$], [$A→B$],
[$0$], [$0$], [$0$], [$0$], [$i_1$],
[$0$], [$1$], [$0$], [$1$], [$i_2$],
[$0$], [$2$], [$0$], [$2$], [$i_3$],
[$1$], [$0$], [$0$], [$1$], [$i_4$],
[$1$], [$1$], [$1$], [$1$], [$i_5$],
[$1$], [$2$], [$1$], [$2$], [$i_6$],
[$2$], [$0$], [$0$], [$2$], [$i_7$],
[$2$], [$1$], [$1$], [$2$], [$i_8$],
[$2$], [$2$], [$2$], [$2$], [$i_9$],
columns: 5, [$A$], [$B$], [$A∧B$], [$A∨B$], [$A→B$], [$0$], [$0$], [$0$], [$0$], [$i_1$], [$0$], [$1$], [$0$], [$1$], [$i_2$], [$0$], [$2$], [$0$], [$2$], [$i_3$], [$1$], [$0$], [$0$], [$1$], [$i_4$], [$1$], [$1$], [$1$], [$1$], [$i_5$], [$1$], [$2$], [$1$], [$2$], [$i_6$], [$2$], [$0$], [$0$], [$2$], [$i_7$], [$2$], [$1$], [$1$], [$2$], [$i_8$], [$2$], [$2$], [$2$], [$2$], [$i_9$],
)
#table(
columns: 2,
[$A$], [¬$A$],
[$0$], [$i_10$],
[$1$], [$i_11$],
[$2$], [$i_12$],
columns: 2, [$A$], [¬$A$], [$0$], [$i_10$], [$1$], [$i_11$], [$2$], [$i_12$],
)
= Model interpreation
@ -400,10 +352,10 @@ 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 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
entities (a set and a mapping):
An interpretation $J$ of the language $L$ consists of the following two entities
(a set and a mapping):
+ 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
@ -416,13 +368,12 @@ entities (a set and a mapping):
- 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
*true formulas* (more precisely − the notion of formulas that are true under
the interpretation $J$).
*true formulas* (more precisely − the notion of formulas that are true under the
interpretation $J$).
*Example.* The above interpretation of the “language about people” put in the
terms of the general definition:
+ $D = {#[br], #[jo], #[pa], #[pe]}$.
+ $#[int]_J (#[Britney])=#[br], #[int]_J (#[John])=#[jo], #[int]_J (#[Paris])=#[pa],
#[int]_J (#[Peter])=#[pe]$.
@ -439,8 +390,8 @@ 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
variables – if any):
+ Truth-values of the formulas: $¬B , B∧C , B∨C , B →C$ [those are not
examples] must be computed. This is done with the truth-values of $B$ and $C$
+ Truth-values of the formulas: $¬B , B∧C , B∨C , B →C$ [those are not examples]
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).
+ The formula $∀x B$ is true under $J$ if and only if $B(c)$ is true under $J$
@ -452,21 +403,19 @@ variables – if any):
*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
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
this interpretation. The notion “a closed formula F is true under the
interpretation J” is now precisely defined.
Similarly, $∀x ∀y(x+ y=y+x)$ is a closed formula that is true under this
interpretation. The notion “a closed formula F is true under the interpretation
J” is now precisely defined.
*Important − non-constructivity!* It may seem that, under an interpretation,
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.
*Important − non-constructivity!* It may seem that, under an interpretation, 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.
=== Example of building of an interpretation
@ -491,55 +440,32 @@ Interpretations of predicate constants are defined by the following truth
tables:
#table(
columns: 3,
[x], [Male(x)], [Female(x)],
[br], [false], [true],
[jo], [true], [false],
[pa], [false], [true],
[pe], [true], [false],
columns: 3, [x], [Male(x)], [Female(x)], [br], [false], [true], [jo], [true], [false], [pa], [false], [true], [pe], [true], [false],
)
#table(
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],
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],
)
== Three kinds of formulas
If one explores some formula F of the language L under various
interpretations, then three situations are possible:
If one explores some formula F of the language L under various interpretations,
then three situations are possible:
+ $F$ is true in all interpretations of the language $L$. Formulas of this kind are
called *logically valid formulas* (LVF, Latv. *LVD*).
+ $F$ is true in all interpretations of the language $L$. Formulas of this kind
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 $L$.
+ 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*).
Formulas that are "not unsatisfiable" (formulas of classes (a) and (b)) are
called, of course, satisfiable formulas: a formula is satisfiable, if it is
true in at least one interpretation [*satisfiable functions* (Latv. *izpildāmas
called, of course, satisfiable formulas: a formula is satisfiable, if it is true
in at least one interpretation [*satisfiable functions* (Latv. *izpildāmas
funkcijas*)].
=== Prooving an F is LVF (Latv. LVD)
First, we should learn to prove that some formula is (if really is!) logically
@ -556,41 +482,36 @@ tests.
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
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,
let us take the domain $D = {a, b}$, and set (in fact, we are using one of two
possibilities):
#table(
columns: 3,
[x], [p(x)], [q(x)],
[b], [false], [true],
[a], [true], [false],
columns: 3, [x], [p(x)], [q(x)], [b], [false], [true], [a], [true], [false],
)
In this interpretation, $p(a)∨q(a) = #[true]$ , $p(b)∨q(b) = #[true]$,
i.e., the premise $∀x( p( x)∨q(x))$ is true. But the formulas$forall p(x),
In this interpretation, $p(a)∨q(a) = #[true]$ , $p(b)∨q(b) = #[true]$, i.e., the
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 $∀x
p(x)∨∀x q(x)$ is false, and $∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)$ is false. We
have built an interpretation, making the formula false. Hence, it is not
logically valid.
p(x)∨∀x q(x)$ is false, and $∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)$ is false. We have
built an interpretation, making the formula false. Hence, it is not logically
valid.
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
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.
== Gödel's Completeness Theorem
*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
$[L_1−L_15,#[MP],#[Gen]]$.
= Tableaux algorithm
== Step 1.
@ -618,9 +538,9 @@ $[L_1−L_15,#[MP],#[Gen]]$.
We will solve the task from the opposite: append to the hypotheses $F_1, dots
F_n$ 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
the set $F_1, ..., F_n, G$, then obtained result will not give an answer
whether $G$ is derivable or not. You should keep this in mind also when the
task has only one formula, e.g., verify, whether formula $(A→B)→((B→C)→(A→C))$
the set $F_1, ..., F_n, G$, then obtained result will not give an answer whether $G$ is
derivable or not. You should keep this in mind also when the task has only one
formula, e.g., verify, whether formula $(A→B)→((B→C)→(A→C))$
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 $F_1, dots,
F_n, ¬G$ 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:
$
(A→B)↔¬A∨B
(A→B)↔¬A∨B
$
Then we apply de Morgan laws to get negations close to the atoms:
$
¬(A∨B)↔¬A∧¬B equiv \
¬(A∧B)↔¬A∨¬B
¬(A∨B)↔¬A∧¬B equiv \
¬(A∧B)↔¬A∨¬B
$
In such way all negations are carried exactly before atoms.
After that we can remove double negations:
In such way all negations are carried exactly before atoms. After that we can
remove double negations:
$
¬¬A↔A
¬¬A↔A
$
Example: $(p→q)→q$.
@ -666,24 +586,24 @@ atoms.
== Step 3.
Next, we should build a tree, vertices of which are formulas. In the root of
the tree we put our formula. Then we have two cases.
Next, we should build a tree, vertices of which are formulas. In the root of the
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
extended with vertices A and B.
+ If vertex is a formula A∨B, then in place of continuation we have branching
into vertex A and vertex B.
+ If vertex is a formula A∨B, then in place of continuation we have branching into
vertex A and vertex B.
In both cases, the initial vertex is marked as processed. Algorithm continues
to process all cases 1 and 2 until all non-atomic vertices have been processed.
In both cases, the initial vertex is marked as processed. Algorithm continues to
process all cases 1 and 2 until all non-atomic vertices have been processed.
== Step 4.
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
(e.g., $A$ and $¬A$), then it is called closed branch. Other branches are
called open branches.
(e.g., $A$ and $¬A$), then it is called closed branch. Other branches are called
open branches.
*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
tree are closed, then formula in the root is unsatisfiable.
formula in the root is satisfiable. And vice versa – if all branches in the tree
are closed, then formula in the root is unsatisfiable.