Theorem sbequ1 1178

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		pm3.4 331	. . . 4 |- ((x = y /\ ph) -> (x = y -> ph))
2		19.8a 1029	. . . 4 |- ((x = y /\ ph) -> E.x(x = y /\ ph))
3	1, 2	jca 288	. . 3 |- ((x = y /\ ph) -> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4		df-sb 1172	. . 3 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
5	3, 4	sylibr 200	. 2 |- ((x = y /\ ph) -> [y / x]ph)
6	5	ex 373	1 |- (x = y -> (ph -> [y / x]ph))

Syntax hints:   -> wi 3   /\ wa 223   = wceq 956  E.wex 980  [wsbc 1170

This theorem is referenced by:  sbequ12 1181  dfsb2 1225  sbequi 1228  sbn 1231  sbi1 1232  hbsb4 1248  sb6rf 1260  mo 1393

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 973

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172

Copyright terms: Public domain