Metamath Proof Explorer


Theorem axdc3lem3

Description: Simple substitution lemma for axdc3 . (Contributed by Mario Carneiro, 27-Jan-2013)

Ref Expression
Hypotheses axdc3lem3.1 AV
axdc3lem3.2 S=s|nωs:sucnAs=CknssuckFsk
axdc3lem3.3 BV
Assertion axdc3lem3 BSmωB:sucmAB=CkmBsuckFBk

Proof

Step Hyp Ref Expression
1 axdc3lem3.1 AV
2 axdc3lem3.2 S=s|nωs:sucnAs=CknssuckFsk
3 axdc3lem3.3 BV
4 2 eleq2i BSBs|nωs:sucnAs=CknssuckFsk
5 feq1 s=Bs:sucnAB:sucnA
6 fveq1 s=Bs=B
7 6 eqeq1d s=Bs=CB=C
8 fveq1 s=Bssuck=Bsuck
9 fveq1 s=Bsk=Bk
10 9 fveq2d s=BFsk=FBk
11 8 10 eleq12d s=BssuckFskBsuckFBk
12 11 ralbidv s=BknssuckFskknBsuckFBk
13 5 7 12 3anbi123d s=Bs:sucnAs=CknssuckFskB:sucnAB=CknBsuckFBk
14 13 rexbidv s=Bnωs:sucnAs=CknssuckFsknωB:sucnAB=CknBsuckFBk
15 3 14 elab Bs|nωs:sucnAs=CknssuckFsknωB:sucnAB=CknBsuckFBk
16 suceq n=msucn=sucm
17 16 feq2d n=mB:sucnAB:sucmA
18 raleq n=mknBsuckFBkkmBsuckFBk
19 17 18 3anbi13d n=mB:sucnAB=CknBsuckFBkB:sucmAB=CkmBsuckFBk
20 19 cbvrexvw nωB:sucnAB=CknBsuckFBkmωB:sucmAB=CkmBsuckFBk
21 4 15 20 3bitri BSmωB:sucmAB=CkmBsuckFBk