Theorem sbequ 1224

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 1223	. 2 |- (x = y -> ([x / z]ph -> [y / z]ph))
2		sbequi 1223	. . 3 |- (y = x -> ([y / z]ph -> [x / z]ph))
3	2	equcoms 1126	. 2 |- (x = y -> ([y / z]ph -> [x / z]ph))
4	1, 3	impbid 514	1 |- (x = y -> ([x / z]ph <-> [y / z]ph))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 953  [wsbc 1166

This theorem is referenced by:  sbco2 1250  sb10f 1337  findes 3150  tfinds 3151  tfindes 3154  nn1suc 5887

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-9 962  ax-10 963  ax-12 965  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-11o 1213

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 978  df-sb 1168

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