Theorem sbequ 1266

Description: An equality theorem for substitution. Used in proof of Theorem 9.7 in [Megill] p. 449 (p. 16 of the preprint).

Assertion

Ref	Expression
sbequ	|- (x = y -> ([x / z]ph <-> [y / z]ph))

Proof of Theorem sbequ

Step	Hyp	Ref	Expression
1		sbequi 1265	. 2 |- (x = y -> ([x / z]ph -> [y / z]ph))
2		sbequi 1265	. . 3 |- (y = x -> ([y / z]ph -> [x / z]ph))
3	2	equcoms 1167	. 2 |- (x = y -> ([y / z]ph -> [x / z]ph))
4	1, 3	impbid 519	1 |- (x = y -> ([x / z]ph <-> [y / z]ph))

Syntax hints:   -> wi 3   <-> wb 144   = wceq 992  [wsbc 1207

This theorem is referenced by:  sbco2 1293  sb10f 1381  tfinds 3212  tfindes 3215  findes 3248  nn1suc 6084

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 998  ax-gen 999  ax-8 1000  ax-9 1001  ax-10 1002  ax-12 1004  ax-4 1009  ax-5o 1011  ax-6o 1014  ax-9o 1159  ax-10o 1177  ax-11o 1255

This theorem depends on definitions:  df-bi 145  df-or 222  df-an 223  df-ex 1017  df-sb 1209

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