Theorem sb4a 1664

Description: A version of sb4 1695 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 1631	. 2 ⊢ ([y / x]∀yφ → ∃x(x = y ∧ ∀yφ))
2		equs5a 1657	. 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 1226  ∃wex 1362  [wsb 1627

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-gen 1318  ax-ie2 1364  ax-11 1378  ax-ial 1409

This theorem depends on definitions:  df-bi 110  df-sb 1628

This theorem is referenced by:  sb6f  1666  hbsb2a  1669