Theorem eqsal 833

Description: A useful equivalence related to substitution.

Hypotheses

Ref	Expression
eqsal.1	⊢ (ψ → ∀xψ)
eqsal.2	⊢ (x = y → (φ ↔ ψ))

Assertion

Ref	Expression
eqsal	⊢ (∀x(x = y → φ) ↔ ψ)

Proof of Theorem eqsal

Step	Hyp	Ref	Expression
1		eqsal.2	. . . . 5 ⊢ (x = y → (φ ↔ ψ))
2		eqsal.1	. . . . . 6 ⊢ (ψ → ∀xψ)
3	2	19.3r 714	. . . . 5 ⊢ (ψ ↔ ∀xψ)
4	1, 3	syl6bb 414	. . . 4 ⊢ (x = y → (φ ↔ ∀xψ))
5	4	pm5.74i 443	. . 3 ⊢ ((x = y → φ) ↔ (x = y → ∀xψ))
6	5	bial 695	. 2 ⊢ (∀x(x = y → φ) ↔ ∀x(x = y → ∀xψ))
7		ax-1 3	. . . . 5 ⊢ (∀xψ → (x = y → ∀xψ))
8	7	a5i 687	. . . 4 ⊢ (∀xψ → ∀x(x = y → ∀xψ))
9	2, 8	syl 12	. . 3 ⊢ (ψ → ∀x(x = y → ∀xψ))
10		ax9 807	. . 3 ⊢ (∀x(x = y → ∀xψ) → ψ)
11	9, 10	impbi 139	. 2 ⊢ (ψ ↔ ∀x(x = y → ∀xψ))
12	6, 11	bitr4 154	1 ⊢ (∀x(x = y → φ) ↔ ψ)

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672   = weq 797

This theorem is referenced by:  eqsex 834  ddelimf2 907  sb6 989

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-gen 677  ax-9 799

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

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