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@ -1,5 +1,38 @@
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#set page(
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margin: (
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left: 1cm,
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right: 1cm,
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top: 1cm,
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bottom: 1cm,
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),
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)
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#set heading(numbering: "1.")
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#show outline.entry.where(
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level: 1
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): it => {
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v(12pt, weak: true)
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strong(it)
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}
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#set enum(numbering: "a)")
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// #show outline.entry.where(
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// level: 1
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// ): it => {
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// v(12pt, weak: true)
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// strong(it)
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// }
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#outline(indent: auto)
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= Logical axiom schemes
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@ -29,10 +62,12 @@ $bold(L_13): F (t)→∃ x F( x)$ (in particular, $F (x)→∃ x F (x)$)
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$bold(L_14): ∀x(G →F (x))→(G→∀x F (x)) $
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$bold(L_15): ∀x(F (x)→G)→(∃z(x+z+1=y).x F (x)→G)$
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$bold(L_15): ∀x(F (x) arrow G) arrow (exists x F (x)→G)$
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= Theorems
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== Prooving directly
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$[L_1, L_2, #[MP]]$:
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+ $((A→B)→(A→C))→(A→(B→C))$. Be careful when assuming hypotheses:
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@ -43,33 +78,36 @@ $[L_1, L_2, #[MP]]$:
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Permutation Law*. Explain the difference between this formula and Theorem
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1.4.3(a): A→(B→C)├ B→(A→C).
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== Theorem 1.5.1 (Deduction Theorem 1 AKA DT1)
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== Deduction theorems
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=== Theorem 1.5.1 (Deduction Theorem 1 AKA DT1)
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If $T$ is a first order theory, and there is a proof of
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$[T, #[MP]]: A_1, A_2, dots, A_n, B├ C$,
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then there is a proof of
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$[L_1, L_2, T, #[MP]]: A_1, A_2, dots, A_n├ B→C$.
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== Theorem 1.5.2 (Deduction Theorem 2 AKA DT2)
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=== Theorem 1.5.2 (Deduction Theorem 2 AKA DT2)
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If there is a proof $[T, #[MP], #[Gen]]: A_1, A_2, dots, A_n, B├ C$,
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where, after B appears in the proof, Generalization is not applied to the
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variables that occur as free in $B$, then there is a proof of
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$[L_1, L_2, L_14, T, #[MP], #[Gen]]: A_1, A_2, dots, A_n├ B→C$.
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== Theorem 2.2.1.
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== Conjunction
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=== Theorem 2.2.1.
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+ (C-introduction): $[L_5, #[MP]]: A, B├ A∧B$;
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+ (C-elimination): $[L_3, L_4, #[MP]]: A∧B ├ A, A∧B ├ B$.
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== Theorem 2.2.2.
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=== Theorem 2.2.2.
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+ $[L_1, L_2, L_5, #[MP]]: (A→(B→C)) ↔ ((A→B)→(A→C))$ (extension of the axiom L_2).
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+ $[L_1-L_4, #[MP]]: (A→B)∧( B→C)→( A→C)$ (another form of the *Law of
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Syllogism*, or *transitivity property of implication*).
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== Theorem 2.2.3 (properties of the conjunction connective).
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=== Theorem 2.2.3 (properties of the conjunction connective).
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$[L_1- L_5, #[MP]]$:
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@ -77,7 +115,7 @@ $[L_1- L_5, #[MP]]$:
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+ $ A∧(B∧C)↔( A∧B)∧C$. Conjunction is associative.
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+ $A∧A↔A$ . Conjunction is idempotent.
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== Theorem 2.2.4 (properties of the equivalence connective).
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=== Theorem 2.2.4 (properties of the equivalence connective).
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$[L_1- L_5, #[MP]]$:
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@ -86,7 +124,9 @@ $[L_1- L_5, #[MP]]$:
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+ $(A↔B)→((B↔C) →((A↔C))$ (transitivity).
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== Theorem 2.3.1
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== Disjunction
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=== Theorem 2.3.1
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+ (D-introduction)$[L_6, L_7, #[MP]]: A├ A∨B; B├ A∨B$;
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+ (D-elimination) If there is a proof $[T, #[MP]]: A_1, A_2, ..., A_n, B├ D$,
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@ -94,25 +134,17 @@ $[L_1- L_5, #[MP]]$:
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L_1, L_2, L_8, #[MP]]: A_1, A_2, dots, A_n, B∨C ├ D$.
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== Theorem 2.3.4.
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Conjunction is distributive to disjunction, and disjunction is distributive to
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conjunction:
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+ $[L_1-L_8, #[MP]]: (A∧B)∨C ↔(A∨C)∧(B∨C)$;
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+ $[L_1-L_8, #[MP]]: (A∨B)∧C ↔(A∧C)∨(B∧C)$ .
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== Theorem 2.3.2
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=== Theorem 2.3.2
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a) $[ L_5, L_6-L_8, #[MP]]: A∨B↔B∨A$ . Disjunction is commutative.
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b) $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨A↔A$ . Disjunction is idempotent.
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== Theorem 2.3.3.
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=== Theorem 2.3.3.
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Disjunction is associative: $[L_1, L_2, L_5, L_6-L_8, #[MP]]: A∨(B∨C)↔(
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A∨B)∨C$.
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== Theorem 2.3.4.
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=== Theorem 2.3.4.
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Conjunction is distributive to disjunction, and disjunction is distributive to
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conjunction:
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@ -120,21 +152,30 @@ conjunction:
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+ $[L_1-L_8, #[MP]]: (A∧B)∨C ↔(A∨C)∧(B∨C)$ .
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+ $[L_1-L_8, #[MP]]: (A∨B)∧C ↔(A∧C)∨(B∧C)$ .
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=== Theorem 2.3.4.
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== Theorem 2.4.1.
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Conjunction is distributive to disjunction, and disjunction is distributive to
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conjunction:
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+ $[L_1-L_8, #[MP]]: (A∧B)∨C ↔(A∨C)∧(B∨C)$;
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+ $[L_1-L_8, #[MP]]: (A∨B)∧C ↔(A∧C)∨(B∧C)$ .
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== Negation -- minimal logic
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=== Theorem 2.4.1.
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(N-elimination) If there is a proof
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$[T, #[MP]]: A_1, A_2, ..., A_n, B├ C$, and a proof $[T, #[MP]]: A_1, A_2, ..., A_n,
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B├ ¬C$, then there is a proof $[T, L_1, L_2, L_9, #[MP]]: A_1, A_2, ..., A_n├ ¬B$.
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== Theorem 2.4.2.
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=== Theorem 2.4.2.
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a) $[L_1, L_2, L_9, #[MP]]: A, ¬B├ ¬(A→B)$. What does it mean?
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b) $[L_1-L_4, L_9, #[MP]]: A∧¬B→¬( A→B)$.
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== Theorem 2.4.3.
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=== Theorem 2.4.3.
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$[L_1, L_2, L_9, #[MP]]: (A→B)→(¬B→¬A)$. What does it mean?
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It's the so-called *Contraposition Law*.
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@ -142,15 +183,11 @@ It's the so-called *Contraposition Law*.
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Note. The following rule form of Contraposition Law is called *Modus Tollens*:
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$[L_1, L_2, L_9, #[MP]]: A→B, ¬B├ ¬A, or, frac(A→B \; ¬B, ¬A)$
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== Theorem 2.4.4.
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=== Theorem 2.4.4.
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$[L_1, L_2, L_9, #[MP]]: A→¬¬A$.
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== Theorem 2.4.5.
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$[L_1, L_2, L_9, #[MP]]: ¬¬¬A↔¬A$.
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== Theorem 2.4.5.
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=== Theorem 2.4.5.
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a) $[L_1, L_2, L_9, #[MP]]: ¬¬¬A↔¬A$.
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b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$. What does it mean? This is a “weak
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@ -158,24 +195,26 @@ b) $[L_1, L_2, L_6, L_7, L_9, #[MP]]: ¬¬( A∨¬A)$. What does it mean? This i
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constructively. The formula $¬¬( A∨¬A)$ can be proved in the constructive logic,
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but $A∨¬A$ can't – as we will see in Section 2.8.
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== Theorem 2.4.9.
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=== Theorem 2.4.9.
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+ $[L_1, L_2, L_8, L_9, #[MP]]: ¬A∨¬B→¬( A∧B)$ . It's the constructive half of the
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so-called *First de Morgan Law*. What does it mean?
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+ $[L_1-L_9, #[MP]]: ¬(A∨B)↔¬A∧¬B$. It's the so-called *Second de Morgan Law*.
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== Theorem 2.5.1.
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== Negation -- constructive logic
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=== Theorem 2.5.1.
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+ $[L_1, L_8, L_10, #[MP]]: ¬A∨B→( A→B)$.
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+ $[L_1, L_2, L_6, #[MP]]: A∨B→(¬A→B) ├¬A→(A→B)$ . It means that the “natural”
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rule $A∨B ;¬ A ├ B$ implies $L_10$!
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== Theorem 2.5.2.
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=== Theorem 2.5.2.
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$[L_1-L_10, #[MP]]$:
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+ $(¬¬A→¬¬B)→¬¬(A→B)$. It's the converse of Theorem 2.4.7(b). Hence, $[L1-L10,
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+ $(¬¬A→¬¬B)→¬¬(A→B)$. It's the converse of Theorem 2.4.7(b). Hence, $[L_1-L_10,
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#[MP]]:├ ¬¬(A→B)↔(¬¬A→¬¬B)$.
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+ $¬¬A→(¬A→A)$. It's the converse of Theorem 2.4.6(a). Hence, [L1-L10,
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#[MP]]: $¬¬A↔(¬A→A)$.
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@ -183,50 +222,59 @@ $[L_1-L_10, #[MP]]$:
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+ $¬¬(¬¬A→A)$. What does it mean? It’s a “weak” form of the Double Negations
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Law – provable in constructive logic.
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== Theorem 2.6.1. (Double Negation Law)
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== Negation -- classical logic
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=== Theorem 2.6.1. (Double Negation Law)
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$[L_1, L_2, L_8, L_10, L_11, #[MP]]: ¬¬A → A$. Hence, $[L_1-L_11, #[MP]]: ¬¬A ↔
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A$.
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== Theorem 2.6.2.
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=== Theorem 2.6.2.
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$[L_8, L_11, #[MP]]: A→B, ¬A→B├ B$. Or, by Deduction Theorem 1, $[L_1, L_2, L_8,
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L_11, #[MP]]: (A→B)→((¬A→B)→B)$.
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== Theorem 2.6.3.
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=== Theorem 2.6.3.
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$[L_1-L_11, #[MP]]: (¬B→¬A)→(A→B)$. Hence, $[L_1-L_11, #[MP]]: (A→B) ↔ (¬B→¬A)$.
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== Theorem 2.6.3.
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=== Theorem 2.6.3.
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_(another one with the same number of because numbering error (it seems like it))_
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$[L_1-L_9, L_11, #[MP]]: ˫ ¬(A∧B)→¬A∨¬B$ . Hence, $[L_1-L_9, L_11, MP]: ˫
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$[L_1-L_9, L_11, #[MP]]: ˫ ¬(A∧B)→¬A∨¬B$ . Hence, $[L_1-L_9, L_11, #[MP]]: ˫
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¬(A∧B)↔¬A∨¬B$ .
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== Theorem 2.6.4.
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=== Theorem 2.6.4.
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$[L_1-L_8, L_11, MP]: (A→B)→¬ A∨B $. Hence, (I-elimination) $[L_1-L_11, #[MP]]:
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$[L_1-L_8, L_11, #[MP]]: (A→B)→¬ A∨B $. Hence, (I-elimination) $[L_1-L_11, #[MP]]:
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(A→B)↔¬ A∨B$.
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== Theorem 2.6.5.
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=== Theorem 2.6.5.
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$[L_1-L_11, MP]: ¬(A→B)→A∧¬B $.
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== Theorem 2.7.1 (Glivenko's Theorem).
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$[L_1-L_11, #[MP]]:├ A$ if and only if $[L1-L10, #[MP]]:├ ¬¬A$.
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$[L_1-L_11, #[MP]]: ¬(A→B)→A∧¬B $.
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== Theorem 2.8.1.
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=== Theorem 2.7.1 (Glivenko's Theorem).
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$[L_1-L_11, #[MP]]:├ A$ if and only if $[L_1-L_10, #[MP]]:├ ¬¬A$.
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== Axiom independence
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=== Theorem 2.8.1.
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The axiom $L_9$: $(A→B)→((A→¬B)→¬A)$ can be proved in $[L_1, L_2, L_8, L_10,
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L_11, #[MP]]$.
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=== Theorem 2.8.2.
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The axiom $L_9$ cannot be proved in $[L_1-L_8, L_10, #[MP]]$.
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== Replacement Theorem 1
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Let us consider three formulas: $B$, $B'$, $C$, where $B$ is a sub-formula of
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@ -503,6 +551,44 @@ tests.
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=== Prooving an F is satisfiable but NOT LVF
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As an example, let us verify that the formula
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$
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∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)
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$
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is not logically valid (p, q are predicate constants). Why it is not? Because
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the truth-values of p(x) and q(x) may behave in such a way that $p(x)∨q(x)$ is
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always true, but neither $forall x p(x)$, nor $forall x q(x)$ is true. Indeed,
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let us take the domain $D = {a, b}$, and set (in fact, we are using one of two
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possibilities):
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#table(
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columns: 3,
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[x], [p(x)], [q(x)],
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[b], [false], [true],
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[a], [true], [false],
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)
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In this interpretation, $p(a)∨q(a) = #[true]$ , $p(b)∨q(b) = #[true]$,
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i.e., the premise $∀x( p( x)∨q(x))$ is true. But the formulas$forall p(x),
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forall q(x)$ both are false. Hence, in this interpretation, the conclusion $∀x
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p(x)∨∀x q(x)$ is false, and $∀x( p( x)∨q( x))→∀x p(x)∨∀x q(x)$ is false. We
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have built an interpretation, making the formula false. Hence, it is not
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logically valid.
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On the other hand, this formula is satisfiable – there is an interpretation
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under which it is true. Indeed, let us take $D={a}$ as the domain of
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interpretation, and let us set $p(a)=q(a)=#[true]$. Then all the formulas
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$
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∀x( p( x)∨q( x)),∀x p(x),∀x q( x)
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$
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become true, and so becomes the entire formula.
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== Gödel's Completeness Theorem
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*Theorem 4.3.1.* In classical predicate logic $[L_1−L_15,#[MP],#[Gen]]$ all
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@ -511,3 +597,91 @@ logically valid formulas can be derived.
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*Theorem 4.3.3.* All formulas that can be derived in classical predicate logic
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$[L_1−L_15,#[MP],#[Gen]]$ are logically valid. In this logic it is not possible
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to derive contradictions, it is consistent.
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=== Gödel’s theorem usage for task solving
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This theorem gives us new method to conclude that some formula $F$ is derivable
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in classical predicate logic: instead of trying to derive $F$ by using axioms,
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rules of inference, deduction theorem, T 2.3.1 and other helping tools, we can
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just prove that $F$ is logically valid (by showing that none of interpretations
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can make it false). If we manage to do so, then we can announce: according to
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Gödel’s theorem, $F$ is derivable in classical predicate logic
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$[L_1−L_15,#[MP],#[Gen]]$.
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= Tableaux algorithm
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== Step 1.
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We will solve the task from the opposite: append to the hypotheses $F_1, dots
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F_n$ negation of formula $G$, and obtain the set $F_1, dots, F_n, ¬G$. When you
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will do homework or test, you shouldn’t forget this, because if you work with
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the set $F_1, ..., F_n, G$, then obtained result will not give an answer
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whether $G$ is derivable or not. You should keep this in mind also when the
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task has only one formula, e.g., verify, whether formula $(A→B)→((B→C)→(A→C))$
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is derivable. Then from the beginning you should append negation in front:
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¬((A→B)→((B→C)→(A→C))) and then work further. Instead of the set $F_1, dots,
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F_n, ¬G$ we can always check one formula $F_1∧...∧F_n∧¬G$. Therefore, our task
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(theoretically) is reducing to the task: given some predicate language formula
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F, verify, whether it is satisfiable or not.
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== Step 2.
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Before applying the algorithm, you first should translate formula to the
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so-called negation normal form. We can use the possibilities provided by
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Substitution theorem. First, implications are replaced with negations and
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disjunctions:
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$
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(A→B)↔¬A∨B
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$
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|
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Then we apply de Morgan laws to get negations close to the atoms:
|
||||
|
||||
$
|
||||
¬(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:
|
||||
|
||||
$
|
||||
¬¬A↔A
|
||||
$
|
||||
|
||||
Example: $(p→q)→q$.
|
||||
|
||||
First get rid of implications: $¬(¬p∨q)∨q$.
|
||||
|
||||
Then apply de Morgan law: $(¬¬p∧¬q)∨q$.
|
||||
|
||||
Then get rid of double negations: $(p∧¬q)∨q$.
|
||||
|
||||
Now we have obtained equivalent formula in negation normal form – formula only
|
||||
has conjunctions and disjunctions, and all negations appear only in front of
|
||||
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.
|
||||
|
||||
+ 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.
|
||||
|
||||
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.
|
||||
|
||||
*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.
|
||||
|
||||
Loading…
Reference in New Issue
Block a user