Theorem equs45f 1680

Description: Two ways of expressing substitution when y is not free in φ. (Contributed by NM, 25-Apr-2008.)

Hypothesis

Ref	Expression
equs45f.1	⊢ (φ → ∀yφ)

Assertion

Ref	Expression
equs45f	⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))

Proof of Theorem equs45f

Step	Hyp	Ref	Expression
1		equs45f.1	. . . . 5 ⊢ (φ → ∀yφ)
2	1	anim2i 324	. . . 4 ⊢ ((x = y ∧ φ) → (x = y ∧ ∀yφ))
3	2	eximi 1488	. . 3 ⊢ (∃x(x = y ∧ φ) → ∃x(x = y ∧ ∀yφ))
4		equs5a 1672	. . 3 ⊢ (∃x(x = y ∧ ∀yφ) → ∀x(x = y → φ))
5	3, 4	syl 14	. 2 ⊢ (∃x(x = y ∧ φ) → ∀x(x = y → φ))
6		equs4 1610	. 2 ⊢ (∀x(x = y → φ) → ∃x(x = y ∧ φ))
7	5, 6	impbii 117	1 ⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1240  ∃wex 1378

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-11 1394  ax-4 1397  ax-i9 1420  ax-ial 1424

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  sb5f  1682