Theorem iffv 6852

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 6834	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐹 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐹‘𝐴))
2		fveq1 6834	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐺 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐺‘𝐴))
3	1, 2	ifsb 4481	1 ⊢ (if(𝜑, 𝐹, 𝐺)‘𝐴) = if(𝜑, (𝐹‘𝐴), (𝐺‘𝐴))

Syntax hints:   = wceq 1542  ifcif 4467  ‘cfv 6493

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3432  df-ss 3907  df-if 4468  df-uni 4852  df-br 5087  df-iota 6449  df-fv 6501

This theorem is referenced by:  decpmatid  22748  pmatcollpwscmatlem1  22767  selvvvval  43035  prjspnfv01  43074