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Metamath Proof Explorer

Theorem fvif

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

Ref Expression
Assertion fvif ⊢ F ⁡ if φ A B = if φ F ⁡ A F ⁡ B

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ if φ A B = A → F ⁡ if φ A B = F ⁡ A
2 fveq2 ⊢ if φ A B = B → F ⁡ if φ A B = F ⁡ B
3 1 2 ifsb ⊢ F ⁡ if φ A B = if φ F ⁡ A F ⁡ B