Theorem iffv 6205

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 6190	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐹 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐹‘𝐴))
2		fveq1 6190	. 2 ⊢ (if(𝜑, 𝐹, 𝐺) = 𝐺 → (if(𝜑, 𝐹, 𝐺)‘𝐴) = (𝐺‘𝐴))
3	1, 2	ifsb 4099	1 ⊢ (if(𝜑, 𝐹, 𝐺)‘𝐴) = if(𝜑, (𝐹‘𝐴), (𝐺‘𝐴))

Syntax hints:   = wceq 1483  ifcif 4086  ‘cfv 5888

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-rex 2918  df-if 4087  df-uni 4437  df-br 4654  df-iota 5851  df-fv 5896

This theorem is referenced by:  decpmatid  20575  pmatcollpwscmatlem1  20594