Theorem sb5rf 1780

Description: Reversed substitution. (Contributed by NM, 3-Feb-2005.) (Proof shortened by Andrew Salmon, 25-May-2011.)

Hypothesis

Ref	Expression
sb5rf.1	⊢ (𝜑 → ∀𝑦𝜑)

Assertion

Ref	Expression
sb5rf	⊢ (𝜑 ↔ ∃𝑦(𝑦 = 𝑥 ∧ [𝑦 / 𝑥]𝜑))

Proof of Theorem sb5rf

Step	Hyp	Ref	Expression
1		sb5rf.1	. . . 4 ⊢ (𝜑 → ∀𝑦𝜑)
2	1	sbid2h 1777	. . 3 ⊢ ([𝑥 / 𝑦][𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		sb1 1696	. . 3 ⊢ ([𝑥 / 𝑦][𝑦 / 𝑥]𝜑 → ∃𝑦(𝑦 = 𝑥 ∧ [𝑦 / 𝑥]𝜑))
4	2, 3	sylbir 133	. 2 ⊢ (𝜑 → ∃𝑦(𝑦 = 𝑥 ∧ [𝑦 / 𝑥]𝜑))
5		stdpc7 1700	. . . 4 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 → 𝜑))
6	5	imp 122	. . 3 ⊢ ((𝑦 = 𝑥 ∧ [𝑦 / 𝑥]𝜑) → 𝜑)
7	1, 6	exlimih 1529	. 2 ⊢ (∃𝑦(𝑦 = 𝑥 ∧ [𝑦 / 𝑥]𝜑) → 𝜑)
8	4, 7	impbii 124	1 ⊢ (𝜑 ↔ ∃𝑦(𝑦 = 𝑥 ∧ [𝑦 / 𝑥]𝜑))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103  ∀wal 1287  ∃wex 1426  [wsb 1692

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-11 1442  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472

This theorem depends on definitions:  df-bi 115  df-sb 1693

This theorem is referenced by:  2sb5rf  1913  sbelx  1921