Theorem iffv 6785

Description: Move a conditional outside of a function. (Contributed by Thierry Arnoux, 28-Sep-2018.)

Assertion

Ref	Expression
iffv	⊢ (if(𝜑, 𝐹, 𝐺)‘𝐴) = if(𝜑, (𝐹‘𝐴), (𝐺‘𝐴))

Proof of Theorem iffv

Step	Hyp	Ref	Expression
1		fveq1 6767	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐹 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐹‘𝐴))
2		fveq1 6767	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐺 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐺‘𝐴))
3	1, 2	ifsb 4477	1 ⊢ (if(𝜑, 𝐹, 𝐺)‘𝐴) = if(𝜑, (𝐹‘𝐴), (𝐺‘𝐴))

Syntax hints:   = wceq 1541  ifcif 4464  ‘cfv 6430

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1544  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-v 3432  df-in 3898  df-ss 3908  df-if 4465  df-uni 4845  df-br 5079  df-iota 6388  df-fv 6438

This theorem is referenced by:  decpmatid  21900  pmatcollpwscmatlem1  21919  prjspnfv01  40441