Theorem ifeq2d 2368

Description: Equality deduction for conditional operator.

Hypothesis

Ref	Expression
ifeq1d.1	|- (ph -> A = B)

Assertion

Ref	Expression
ifeq2d	|- (ph -> if(ps, C, A) = if(ps, C, B))

Proof of Theorem ifeq2d

Step	Hyp	Ref	Expression
1		ifeq1d.1	. 2 |- (ph -> A = B)
2		ifeq2 2363	. 2 |- (A = B -> if(ps, C, A) = if(ps, C, B))
3	1, 2	syl 10	1 |- (ph -> if(ps, C, A) = if(ps, C, B))

Syntax hints:   -> wi 3   = wceq 955  ifcif 2359

This theorem is referenced by:  rdgeq1 3931  oev 4150  unxpdomlem 4830  expvalt 6520  spwval2 8636

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 980  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-if 2360

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