Theorem sbequ 1229

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 1228	. 2 |- (x = y -> ([x / z]ph -> [y / z]ph))
2		sbequi 1228	. . 3 |- (y = x -> ([y / z]ph -> [x / z]ph))
3	2	equcoms 1130	. 2 |- (x = y -> ([y / z]ph -> [x / z]ph))
4	1, 3	impbid 516	1 |- (x = y -> ([x / z]ph <-> [y / z]ph))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 956  [wsbc 1170

This theorem is referenced by:  sbco2 1255  sb10f 1342  findes 3160  tfinds 3161  tfindes 3164  nn1suc 5927

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-9 965  ax-10 966  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172

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