Theorem sbi1 884

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 864	. . . . 5 ⊢ (x = y → ([y / x](φ → ψ) → (φ → ψ)))
2		sbequ2 864	. . . . 5 ⊢ (x = y → ([y / x]φ → φ))
3	1, 2	syl5d 53	. . . 4 ⊢ (x = y → ([y / x](φ → ψ) → ([y / x]φ → ψ)))
4		sbequ1 863	. . . 4 ⊢ (x = y → (ψ → [y / x]ψ))
5	3, 4	syl6d 54	. . 3 ⊢ (x = y → ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ)))
6	5	a4s 682	. 2 ⊢ (∀x x = y → ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ)))
7		sb4 861	. . . 4 ⊢ (¬ ∀x x = y → ([y / x](φ → ψ) → ∀x(x = y → (φ → ψ))))
8		ax-2 4	. . . . . 6 ⊢ ((x = y → (φ → ψ)) → ((x = y → φ) → (x = y → ψ)))
9	8	19.20ii 692	. . . . 5 ⊢ (∀x(x = y → (φ → ψ)) → (∀x(x = y → φ) → ∀x(x = y → ψ)))
10		sb2 859	. . . . 5 ⊢ (∀x(x = y → ψ) → [y / x]ψ)
11	9, 10	syl6 23	. . . 4 ⊢ (∀x(x = y → (φ → ψ)) → (∀x(x = y → φ) → [y / x]ψ))
12	7, 11	syl6 23	. . 3 ⊢ (¬ ∀x x = y → ([y / x](φ → ψ) → (∀x(x = y → φ) → [y / x]ψ)))
13		sb4 861	. . 3 ⊢ (¬ ∀x x = y → ([y / x]φ → ∀x(x = y → φ)))
14	12, 13	syl5d 53	. 2 ⊢ (¬ ∀x x = y → ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ)))
15	6, 14	pm2.61i 110	1 ⊢ ([y / x](φ → ψ) → ([y / x]φ → [y / x]ψ))

Syntax hints:  ¬ wn 1   → wi 2  ∀wal 672   = weq 797  [wsb 852

This theorem is referenced by:  sbim 886

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

metamath.org