Theorem sbi1 1230

Description: Removal of implication from substitution.

Assertion

Ref	Expression
sbi1	⊢ ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ))

Proof of Theorem sbi1

Step	Hyp	Ref	Expression
1		sbequ2 1177	. . . . 5 ⊢ (x = y → ([y / x](φ → ψ) → (φ → ψ)))
2		sbequ2 1177	. . . . 5 ⊢ (x = y → ([y / x]φ → φ))
3	1, 2	syl5d 55	. . . 4 ⊢ (x = y → ([y / x](φ → ψ) → ([y / x]φ → ψ)))
4		sbequ1 1176	. . . 4 ⊢ (x = y → (ψ → [y / x]ψ))
5	3, 4	syl6d 56	. . 3 ⊢ (x = y → ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ)))
6	5	a4s 982	. 2 ⊢ (∀x x = y → ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ)))
7		sb4 1221	. . . 4 ⊢ (¬ ∀x x = y → ([y / x](φ → ψ) → ∀x(x = y → (φ → ψ))))
8		ax-2 5	. . . . . 6 ⊢ ((x = y → (φ → ψ)) → ((x = y → φ) → (x = y → ψ)))
9	8	19.20ii 993	. . . . 5 ⊢ (∀x(x = y → (φ → ψ)) → (∀x(x = y → φ) → ∀x(x = y → ψ)))
10		sb2 1175	. . . . 5 ⊢ (∀x(x = y → ψ) → [y / x]ψ)
11	9, 10	syl6 22	. . . 4 ⊢ (∀x(x = y → (φ → ψ)) → (∀x(x = y → φ) → [y / x]ψ))
12	7, 11	syl6 22	. . 3 ⊢ (¬ ∀x x = y → ([y / x](φ → ψ) → (∀x(x = y → φ) → [y / x]ψ)))
13		sb4 1221	. . 3 ⊢ (¬ ∀x x = y → ([y / x]φ → ∀x(x = y → φ)))
14	12, 13	syl5d 55	. 2 ⊢ (¬ ∀x x = y → ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ)))
15	6, 14	pm2.61i 126	1 ⊢ ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ))

Syntax hints:  ¬ wn 2   → wi 3  ∀wal 952  [wsbc 1168

This theorem is referenced by:  sbim 1232

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-10 964  ax-12 966  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170

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