Theorem equs45f 1989

Description: Two ways of expressing substitution when y is not free in φ. (Contributed by NM, 25-Apr-2008.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
equs45f.1	⊢ Ⅎyφ

Assertion

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

Proof of Theorem equs45f

Step	Hyp	Ref	Expression
1		equs45f.1	. . . . . 6 ⊢ Ⅎyφ
2	1	nfri 1762	. . . . 5 ⊢ (φ → ∀yφ)
3	2	anim2i 552	. . . 4 ⊢ ((x = y ∧ φ) → (x = y ∧ ∀yφ))
4	3	eximi 1576	. . 3 ⊢ (∃x(x = y ∧ φ) → ∃x(x = y ∧ ∀yφ))
5		equs5a 1887	. . 3 ⊢ (∃x(x = y ∧ ∀yφ) → ∀x(x = y → φ))
6	4, 5	syl 15	. 2 ⊢ (∃x(x = y ∧ φ) → ∀x(x = y → φ))
7		equs4 1959	. 2 ⊢ (∀x(x = y → φ) → ∃x(x = y ∧ φ))
8	6, 7	impbii 180	1 ⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541  Ⅎwnf 1544

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is referenced by:  sb5f  2040