Metamath Proof Explorer


Theorem fvif

Description: Move a conditional outside of a function. (Contributed by Jeff Madsen, 2-Sep-2009)

Ref Expression
Assertion fvif FifφAB=ifφFAFB

Proof

Step Hyp Ref Expression
1 fveq2 ifφAB=AFifφAB=FA
2 fveq2 ifφAB=BFifφAB=FB
3 1 2 ifsb FifφAB=ifφFAFB