Metamath Proof Explorer


Theorem offveqb

Description: Equivalent expressions for equality with a function operation. (Contributed by NM, 9-Oct-2014) (Proof shortened by Mario Carneiro, 5-Dec-2016)

Ref Expression
Hypotheses offveq.1 φAV
offveq.2 φFFnA
offveq.3 φGFnA
offveq.4 φHFnA
offveq.5 φxAFx=B
offveq.6 φxAGx=C
Assertion offveqb φH=FRfGxAHx=BRC

Proof

Step Hyp Ref Expression
1 offveq.1 φAV
2 offveq.2 φFFnA
3 offveq.3 φGFnA
4 offveq.4 φHFnA
5 offveq.5 φxAFx=B
6 offveq.6 φxAGx=C
7 dffn5 HFnAH=xAHx
8 4 7 sylib φH=xAHx
9 inidm AA=A
10 2 3 1 1 9 5 6 offval φFRfG=xABRC
11 8 10 eqeq12d φH=FRfGxAHx=xABRC
12 fvexd φHxV
13 12 ralrimivw φxAHxV
14 mpteqb xAHxVxAHx=xABRCxAHx=BRC
15 13 14 syl φxAHx=xABRCxAHx=BRC
16 11 15 bitrd φH=FRfGxAHx=BRC