Metamath Proof Explorer


Theorem ssmclslem

Description: Lemma for ssmcls . (Contributed by Mario Carneiro, 18-Jul-2016)

Ref Expression
Hypotheses mclsval.d D=mDVT
mclsval.e E=mExT
mclsval.c C=mClsT
mclsval.1 φTmFS
mclsval.2 φKD
mclsval.3 φBE
ssmclslem.h H=mVHT
Assertion ssmclslem φBranHKCB

Proof

Step Hyp Ref Expression
1 mclsval.d D=mDVT
2 mclsval.e E=mExT
3 mclsval.c C=mClsT
4 mclsval.1 φTmFS
5 mclsval.2 φKD
6 mclsval.3 φBE
7 ssmclslem.h H=mVHT
8 simpl BranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspcBranHc
9 8 a1i φBranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspcBranHc
10 9 alrimiv φcBranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspcBranHc
11 ssintab BranHc|BranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspccBranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspcBranHc
12 10 11 sylibr φBranHc|BranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspc
13 eqid mAxT=mAxT
14 eqid mSubstT=mSubstT
15 eqid mVarsT=mVarsT
16 1 2 3 4 5 6 7 13 14 15 mclsval φKCB=c|BranHcmopmopmAxTsranmSubstTsoranHcxyxmymVarsTsHx×mVarsTsHyKspc
17 12 16 sseqtrrd φBranHKCB