Theorem sbequ2 864

Description: An equality theorem for substitution.

Assertion

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

Proof of Theorem sbequ2

Step	Hyp	Ref	Expression
1		pm3.26 256	. . 3 |- (((x = y -> ph) /\ E.x(x = y /\ ph)) -> (x = y -> ph))
2	1	com12 13	. 2 |- (x = y -> (((x = y -> ph) /\ E.x(x = y /\ ph)) -> ph))
3		df-sb 853	. 2 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4	2, 3	syl5ib 181	1 |- (x = y -> ([y / x]ph -> ph))

Syntax hints:   -> wi 2   /\ wa 196  E.wex 678   = weq 797  [wsb 852

This theorem is referenced by:  sbequ12 865  sb6y 872  hbsb3 875  sbequi 876  sbn1 880  sbi1 884  hbsb4 905  mo 1020  mopick 1054

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198  df-sb 853

metamath.org