Metamath Proof Explorer


Theorem sbcel12

Description: Distribute proper substitution through a membership relation. (Contributed by NM, 10-Nov-2005) (Revised by NM, 18-Aug-2018)

Ref Expression
Assertion sbcel12 [˙A/x]˙BCA/xBA/xC

Proof

Step Hyp Ref Expression
1 dfsbcq2 z=AzxBC[˙A/x]˙BC
2 dfsbcq2 z=AzxyB[˙A/x]˙yB
3 2 abbidv z=Ay|zxyB=y|[˙A/x]˙yB
4 dfsbcq2 z=AzxyC[˙A/x]˙yC
5 4 abbidv z=Ay|zxyC=y|[˙A/x]˙yC
6 3 5 eleq12d z=Ay|zxyBy|zxyCy|[˙A/x]˙yBy|[˙A/x]˙yC
7 nfs1v xzxyB
8 7 nfab _xy|zxyB
9 nfs1v xzxyC
10 9 nfab _xy|zxyC
11 8 10 nfel xy|zxyBy|zxyC
12 sbab x=zB=y|zxyB
13 sbab x=zC=y|zxyC
14 12 13 eleq12d x=zBCy|zxyBy|zxyC
15 11 14 sbiev zxBCy|zxyBy|zxyC
16 1 6 15 vtoclbg AV[˙A/x]˙BCy|[˙A/x]˙yBy|[˙A/x]˙yC
17 df-csb A/xB=y|[˙A/x]˙yB
18 df-csb A/xC=y|[˙A/x]˙yC
19 17 18 eleq12i A/xBA/xCy|[˙A/x]˙yBy|[˙A/x]˙yC
20 16 19 bitr4di AV[˙A/x]˙BCA/xBA/xC
21 sbcex [˙A/x]˙BCAV
22 21 con3i ¬AV¬[˙A/x]˙BC
23 noel ¬A/xB
24 csbprc ¬AVA/xC=
25 24 eleq2d ¬AVA/xBA/xCA/xB
26 23 25 mtbiri ¬AV¬A/xBA/xC
27 22 26 2falsed ¬AV[˙A/x]˙BCA/xBA/xC
28 20 27 pm2.61i [˙A/x]˙BCA/xBA/xC