Theorem iffv 6937

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 6919	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐹 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐹‘𝐴))
2		fveq1 6919	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐺 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐺‘𝐴))
3	1, 2	ifsb 4561	1 ⊢ (if(𝜑, 𝐹, 𝐺)‘𝐴) = if(𝜑, (𝐹‘𝐴), (𝐺‘𝐴))

Syntax hints:   = wceq 1537  ifcif 4548  ‘cfv 6573

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-ss 3993  df-if 4549  df-uni 4932  df-br 5167  df-iota 6525  df-fv 6581

This theorem is referenced by:  decpmatid  22797  pmatcollpwscmatlem1  22816  selvvvval  42540  prjspnfv01  42579