Metamath Proof Explorer


Theorem ofsubeq0

Description: Function analogue of subeq0 . (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Assertion ofsubeq0 AVF:AG:AFfG=A×0F=G

Proof

Step Hyp Ref Expression
1 simp2 AVF:AG:AF:A
2 1 ffnd AVF:AG:AFFnA
3 simp3 AVF:AG:AG:A
4 3 ffnd AVF:AG:AGFnA
5 simp1 AVF:AG:AAV
6 inidm AA=A
7 eqidd AVF:AG:AxAFx=Fx
8 eqidd AVF:AG:AxAGx=Gx
9 2 4 5 5 6 7 8 ofval AVF:AG:AxAFfGx=FxGx
10 c0ex 0V
11 10 fvconst2 xAA×0x=0
12 11 adantl AVF:AG:AxAA×0x=0
13 9 12 eqeq12d AVF:AG:AxAFfGx=A×0xFxGx=0
14 1 ffvelcdmda AVF:AG:AxAFx
15 3 ffvelcdmda AVF:AG:AxAGx
16 14 15 subeq0ad AVF:AG:AxAFxGx=0Fx=Gx
17 13 16 bitrd AVF:AG:AxAFfGx=A×0xFx=Gx
18 17 ralbidva AVF:AG:AxAFfGx=A×0xxAFx=Gx
19 2 4 5 5 6 offn AVF:AG:AFfGFnA
20 10 fconst A×0:A0
21 ffn A×0:A0A×0FnA
22 20 21 ax-mp A×0FnA
23 eqfnfv FfGFnAA×0FnAFfG=A×0xAFfGx=A×0x
24 19 22 23 sylancl AVF:AG:AFfG=A×0xAFfGx=A×0x
25 eqfnfv FFnAGFnAF=GxAFx=Gx
26 2 4 25 syl2anc AVF:AG:AF=GxAFx=Gx
27 18 24 26 3bitr4d AVF:AG:AFfG=A×0F=G