Theorem iffv 6884

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 6866	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐹 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐹‘𝐴))
2		fveq1 6866	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐺 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐺‘𝐴))
3	1, 2	ifsb 4494	1 ⊢ (if(𝜑, 𝐹, 𝐺)‘𝐴) = if(𝜑, (𝐹‘𝐴), (𝐺‘𝐴))

Syntax hints:   = wceq 1560  ifcif 4480  ‘cfv 6521

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1563  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3456  df-ss 3921  df-if 4481  df-uni 4866  df-br 5101  df-iota 6477  df-fv 6529

This theorem is referenced by:  selvvvval  22195  decpmatid  22830  pmatcollpwscmatlem1  22849  prjspnfv01  43206