Metamath Proof Explorer


Theorem f1ocpbl

Description: An injection is compatible with any operations on the base set. (Contributed by Mario Carneiro, 24-Feb-2015)

Ref Expression
Hypothesis f1ocpbl.f φ F : V 1-1 onto X
Assertion f1ocpbl φ A V B V C V D V F A = F C F B = F D F A + ˙ B = F C + ˙ D

Proof

Step Hyp Ref Expression
1 f1ocpbl.f φ F : V 1-1 onto X
2 1 f1ocpbllem φ A V B V C V D V F A = F C F B = F D A = C B = D
3 oveq12 A = C B = D A + ˙ B = C + ˙ D
4 3 fveq2d A = C B = D F A + ˙ B = F C + ˙ D
5 2 4 syl6bi φ A V B V C V D V F A = F C F B = F D F A + ˙ B = F C + ˙ D