Theorem sb4a 1682

Description: A version of sb4 1713 that doesn't require a distinctor antecedent. (Contributed by NM, 2-Feb-2007.)

Assertion

Ref	Expression
sb4a	⊢ ([𝑦 / 𝑥]∀𝑦𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))

Proof of Theorem sb4a

Step	Hyp	Ref	Expression
1		sb1 1649	. 2 ⊢ ([𝑦 / 𝑥]∀𝑦𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ ∀𝑦𝜑))
2		equs5a 1675	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ ∀𝑦𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	1, 2	syl 14	1 ⊢ ([𝑦 / 𝑥]∀𝑦𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ∧ wa 97  ∀wal 1241  ∃wex 1381  [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-gen 1338  ax-ie2 1383  ax-11 1397  ax-ial 1427

This theorem depends on definitions:  df-bi 110  df-sb 1646

This theorem is referenced by:  sb6f  1684  hbsb2a  1687