Theorem eqsex 834

Description: A useful equivalence related to substitution.

Hypotheses

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

Assertion

Ref	Expression
eqsex	⊢ (∃x(x = y ∧ φ) ↔ ψ)

Proof of Theorem eqsex

Step	Hyp	Ref	Expression
1		exnal 721	. 2 ⊢ (∃x ¬ (x = y → ¬ φ) ↔ ¬ ∀x(x = y → ¬ φ))
2		df-an 198	. . 3 ⊢ ((x = y ∧ φ) ↔ ¬ (x = y → ¬ φ))
3	2	biex 733	. 2 ⊢ (∃x(x = y ∧ φ) ↔ ∃x ¬ (x = y → ¬ φ))
4		eqsex.1	. . . . 5 ⊢ (ψ → ∀xψ)
5	4	hbne 699	. . . 4 ⊢ (¬ ψ → ∀x ¬ ψ)
6		eqsex.2	. . . . 5 ⊢ (x = y → (φ ↔ ψ))
7	6	negbid 463	. . . 4 ⊢ (x = y → (¬ φ ↔ ¬ ψ))
8	5, 7	eqsal 833	. . 3 ⊢ (∀x(x = y → ¬ φ) ↔ ¬ ψ)
9	8	bicon2i 194	. 2 ⊢ (ψ ↔ ¬ ∀x(x = y → ¬ φ))
10	1, 3, 9	3bitr4 158	1 ⊢ (∃x(x = y ∧ φ) ↔ ψ)

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672  ∃wex 678   = weq 797

This theorem is referenced by:  cleljust 985  sb5 988

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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