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substitution theorem for propositional
logic∗
CWoo†
2013-03-22 3:57:08
In this entry, we will prove the substitution theorem for propositional logic
based on the axiom system found here. Besides the deduction theorem, below
are some additional results we will need to prove the theorem:
1. If ∆ ` A → B and Γ ` B → C, then ∆, Γ ` A → C.
2. ∆ ` A and ∆ ` B iff ∆ ` A ∧ B.
3. ` A ↔ A.
4. ` A ↔ ¬¬A (law of double negation).
5. ⊥→ A (ex falso quodlibet)
6. ∆ ` A implies ∆ ` B iff ∆ ` A → B.
The proofs of these results can be found here.
Theorem 1. (Substitution Theorem) Suppose p1 , . . . , pm are all the propositional variables, not necessarily distinct, that occur in order in A, and if
B1 , . . . , Bm , C1 , . . . , Cm are wff ’s such that ` Bi ↔ Ci , then
` A[B1 /p1 , . . . , Bm /pm ] ↔ A[C1 /p1 , . . . , Cm /pm ]
where A[X1 /p1 , . . . , Xm /pm ] is the wff obtained from A by replacing pi by the
wff Xi via simultaneous substitution.
Proof. We do induction on the number n of → in wff A.
If n = 0, A is either a propositional variable, say p, or ⊥, which respectively
means that A[B/p] ↔ A[C/p] is either B ↔ C or ⊥↔⊥. The former is the
assumption and the latter is a theorem.
∗ hSubstitutionTheoremForPropositionalLogici created: h2013-03-2i by: hCWooi version:
h42534i Privacy setting: h1i hResulti h03B05i
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Suppose now A has n + 1 occurrences of →. We may write A as X → Y
uniquely by unique readability. Also, both X and Y have at most n occurrences
of →.
Let A1 be A[B1 /p1 , . . . , Bm /pm ] and A2 be A[C1 /p1 , . . . , Cm /pm ]. Then A1
is X1 → Y1 and X2 → Y2 , where X1 is X[B1 /p1 , . . . , Bk /pk ], Y1 is Y [Bk+1 /pk+1 , . . . , Bm /pm ],
X2 is X[C1 /p1 , . . . , Ck /pk ], and Y2 is Y [Ck+1 /pk+1 , . . . , Cm /pm ].
Then
` X1 ↔ X2
(1)
by induction
` X1 → X2 and ` X2 → X1
(2)
by 2 above
` Y1 ↔ Y2
(3)
by induction
` Y1 → Y2 and ` Y2 → Y1
(4)
by 2 above
since A1 is X1 → Y1
A1 ` X1 → Y1
(5)
by applying 1 to ` X2 → X1 and (5)
A1 ` X2 → Y1
(6)
by applying 1 to (6) and ` Y1 → Y2
A1 ` X2 → Y2
(7)
by the deduction theorem
` A1 → A2
(8)
by a similar reasoning as above
` A2 → A1
(9)
by applying 2 to (8) and (9)
` A1 ↔ A2
(10)
As a corollary, we have
Corollary 1. If ` B ↔ C, then ` A[B/s(p)] ↔ A[C/s(p)], where p is a
propositional variable that occurs in A, s(p) is a set of positions of occurrences
of p in A, and the wff A[X/s(p)] is obtained by replacing all p that occur in the
positions in s(p) in A by wff X.
Proof. For any propositional variable q not being replaced, use the corresponding theorem ` q ↔ q, and then apply the substitution theorem.
Remark. What about ` B[A/p] ↔ C[A/p], given ` B ↔ C? Here, B[A/p]
and C[A/p] are wff’s obtained by uniform substitution of p (all occurrences of
p) in B and C respectively. Since B[A/p] ↔ C[A/p] is just (B ↔ C)[A/p], an
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instance of the schema B ↔ C by assumption, the result follows directly if we
assume B ↔ C is a theorem schema.
Using the substitution theorem, we can easily derive more theorem schemas,
such as
7. (A → B) ↔ (¬B → ¬A) (Law of Contraposition)
8. A → (¬B → ¬(A → B))
9. ((A → B) → A) → A (Peirce’s Law)
Proof.
7. Since (¬B → ¬A) → (A → B) is already a theorem schema,
we only need to show ` (A → B) → (¬B → ¬A). By law of double
negation (4 above) and the substitution theorem, it is enough to show that
` (¬¬A → ¬¬B) → (¬B → ¬A). But this is just an instance of an axiom
schema. Combining the two schemas, we get ` (A → B) ↔ (¬B → ¬A).
8. First, observe that A, A → B ` B by modus ponens. Since ` B ↔ ¬¬B,
we have A, A → B ` ¬¬B by the substitution theorem. So A, A →
B, ¬B `⊥ by the deduction theorem, and A, ¬B ` (A → B) →⊥ by the
deduction theorem again. Apply the deduction two more times, we get
` A → (¬B → ¬(A → B)).
9. To show ` ((A → B) → A) → A, it is enough to show ` ¬A → ¬((A →
B) → A) by 7 and modus ponens, or ¬A ` ¬((A → B) → A) by the
deduction theorem. Now, since ` X ∧ Y ↔ ¬(X → ¬Y ) (as they are
the same thing, and because C ↔ C is a theorem schema), by the law
of double negation and the substitution theorem, ` X ∧ ¬Y ↔ ¬(X →
Y ), and we have ` (A → B) ∧ ¬A ↔ ¬((A → B) → A). So to show
¬A ` ¬((A → B) → A), it is enough to show ¬A ` (A → B) ∧ ¬A,
which is enough to show that ¬A ` A → B and ¬A ` ¬A, according to
a meta-theorem found here. To show ¬A ` A → B, it is enough to show
{¬A, A} ` B, and A, ¬A, ⊥, ⊥→ B, B is such a deduction. The second
statement ¬A ` ¬A is clear.
As an application, we prove the following useful meta-theorems of propositional logic:
Proposition 1. There is a wff A such that ∆ ` A and ∆ ` ¬A iff ∆ `⊥
Proof. Assume the former. Let E1 be a deduction of A from ∆ and E2 a deduction of ¬A from ∆, then
E1 , E2 , ⊥
is a deduction of ⊥ from ∆. Conversely, assume the later. Pick any wff A (if
necessary, pick ⊥). Then ⊥→ A by ex falso quodlibet. By modus ponens, we
have ∆ ` A. Similarly, ∆ ` ¬A.
Proposition 2. If ∆, A ` B and ∆, ¬A ` B, then ∆ ` B
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Proof. By assumption, we have ∆ ` A → B and ∆ ` ¬A → B. Using modus
ponens and the theorem schema (A → B) → (¬B → ¬A), we have ∆ ` ¬B →
¬A, or
∆, ¬B ` ¬A.
Similarly, ∆ ` ¬B → ¬¬A. By the law of double negation and the substitution
theorem, we have ∆ ` ¬B → A, or
∆, ¬B ` A.
By the previous proposition, ∆, ¬B `⊥, or ∆ ` ¬¬B. Applying the substitution
theorem and the law of double negation, we have
∆ ` B.
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