Theorem sbequ2 1699

Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbequ2	|- ( x = y -> ( [ y / x ] ph -> ph ) )

Proof of Theorem sbequ2

Step	Hyp	Ref	Expression
1		df-sb 1693	. 2 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
2		simpl 107	. . 3 |- ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> ( x = y -> ph ) )
3	2	com12 30	. 2 |- ( x = y -> ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> ph ) )
4	1, 3	syl5bi 150	1 |- ( x = y -> ( [ y / x ] ph -> ph ) )

Syntax hints:   -> wi 4   /\ wa 102   E.wex 1426   [wsb 1692

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104

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

This theorem is referenced by:  stdpc7  1700  sbequ12  1701  sbequi  1767  mo23  1989  mopick  2026