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some meta-theorems of propositional
logic∗
CWoo†
2013-03-22 4:00:01
Based on the axiom system in this entry, we will prove some meta-theorems
of propositional logic. In the discussion below, ∆ and Γ are sets of well-formed
formulas (wff’s), and A, B, C, . . . are wff’s.
1. (Deduction Theorem) ∆, A ` B iff ∆ ` A → B.
2. (Proof by Contradiction) ∆, A `⊥ iff ∆ ` ¬A.
3. (Proof by Contrapositive) ∆, A ` ¬B iff ∆, B ` ¬A.
4. (Law of Syllogism) If ∆ ` A → B and Γ ` B → C, then ∆, Γ ` A → C.
5. ∆ ` A and ∆ ` B iff ∆ ` A ∧ B.
6. ∆ ` A ↔ B iff ∆, A ` B and ∆, B ` A.
7. If ∆ ` A ↔ B, then ∆ ` B ↔ A.
8. If ∆ ` A ↔ B and ∆ ` B ↔ C, then ∆ ` A ↔ C.
9. ∆ ` A ∧ B → C iff ∆ ` A → (B → C).
10. ∆ ` A implies ∆ ` B iff ∆ ` A → B. This is a useful restatement of the
deduction theorem.
11. (Substitution Theorem) If ` Bi ↔ Ci , then ` A[B/p] ↔ A[C/p].
12. ∆ `⊥ iff there is a wff A such that ∆ ` A and ∆ ` ¬A.
13. If ∆, A ` B and ∆, ¬A ` B, then ∆ ` B.
Remark. The theorem schema A → ¬¬A is used in the proofs below.
∗ hSomeMetatheoremsOfPropositionalLogici
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Proof. The first three are proved here, and the last three are proved here. We
will prove the rest here, some of which relies on the deduction theorem.
4. From ∆ ` A → B, by the deduction theorem, we have ∆, A ` B. Let E1
be a deduction of B from ∆ ∪ {A}, and E2 be a deduction of B → C from
Γ, then
E1 , E2 , C
is a deduction of C from ∆∪{A}∪Γ, so ∆, A, Γ ` C, and by the deduction
theorem again, we get ∆, Γ ` A → C.
5. (⇒). Since A ∧ B is ¬(A → ¬B), by the deduction theorem, it is enough
to show ∆, A → ¬B `⊥. Suppose E1 is a deduction of A from ∆ and E2
is a deduction of B from ∆, then
E1 , E2 , A → ¬B, ¬B, ⊥
is a deduction of ⊥ from ∆ ∪ {A → ¬B}.
(⇐). We first show that ∆ ` B. Now, ¬B → (A → ¬B) is an axiom
and ` (A → ¬B) → ¬¬(A → ¬B) is a theorem, ` ¬B → ¬¬(A → ¬B),
so that by modus ponens, ` ¬(A → ¬B) → B, using axiom schema
(¬C → ¬D) → (D → C). Since by assumption ∆ ` ¬(A → ¬B), by
modus ponens again, we get ∆ ` B.
We next show that ∆ ` A. From the deduction A, A →⊥, ⊥, we have
A, ¬A `⊥, so certainly ∆, ¬A, A, B `⊥. By three applications of the
deduction theorem, we get ∆ ` ¬A → (A → ¬B). By theorem (A →
¬B) → ¬¬(A → ¬B), ∆ ` ¬A → ¬¬(A → ¬B). By axiom schema
(¬C → ¬D) → (D → C) and modus ponens, we get ∆ ` ¬(A → ¬B) →
A. Since ∆ ` ¬A → ¬B by assumption, ∆ → A as a result.
6. ∆ ` A ↔ B iff ∆ ` A → B and ∆ ` B → A iff ∆, A ` B and ∆, B ` A.
7. Apply 6 to ∆ ` A → B and ∆ ` B → A.
8. Apply 5 and 6.
9. Since ∆, A ` B → A ∧ B by the theorem schema ` A → (B → A ∧ B),
together with ∆ ` A ∧ B → C, we have ∆, A ` B → C by law of
syllogism, or equivalently ∆ ` A → (B → C), by the deduction theorem.
Conversely, ∆, A ` B → C and theorem schema A ∧ B → B result in
∆, A ` A ∧ B → C by law of syllogism. So ∆ ` A → (A ∧ B → C) by the
deduction theorem. But A ∧ B → A is a theorem schema, ∆ ` A ∧ B →
(A ∧ B → C), and therefore ∆ ` A ∧ B → C by the theorem schema
(X → (X → Y )) ↔ (X → Y ).
10. Assume the former. Then a deduction of B from ∆ may or may not
contain A. In either case, ∆, A ` B, so ∆ ` A → B by the deduction
theorem. Next, assume the later. Let E1 be a deduction of A → B from
∆. Then if E2 is a deduction of A from ∆, then E1 , E2 , B is a deduction of
B from ∆, and therefore ∆ ` B.
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To see the last meta-theorem implies the deduction theorem, assume ∆, A ` B.
Suppose ∆ ` A. Let E1 be a deduction of A from ∆, and E2 a deduction of B
from ∆ ∪ {A}. Then E1 , E2 is a deduction of B from ∆. So ∆ ` B. As a result
∆A → B by the statement of the meta-theorem.
Remark. Meta-theorems 7 and 8, together with the theorem schema A ↔
A, show that ↔ defines an equivalence relation on the set of all wff’s of propositional logic. Formally, for any two wff’s A, B, if we define A ∼ B iff ` A ↔ B,
then ∼ is an equivalence relation.
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