Theorem sbequ1 863

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ1	|- (x = y -> (ph -> [y / x]ph))

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 266	. . . 4 |- ((x = y /\ ph) -> (x = y -> ph))
2		19.8a 712	. . . 4 |- ((x = y /\ ph) -> E.x(x = y /\ ph))
3	1, 2	jca 236	. . 3 |- ((x = y /\ ph) -> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4		df-sb 853	. . 3 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
5	3, 4	sylibr 175	. 2 |- ((x = y /\ ph) -> [y / x]ph)
6	5	exp 291	1 |- (x = y -> (ph -> [y / x]ph))

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

This theorem is referenced by:  sbequ12 865  hbsb3 875  sbequi 876  sbn2 881  sbi1 884  hbsb4 905  sb5f1 917  mo 1020

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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