Metamath Proof Explorer


Theorem sbccom2lem

Description: Lemma for sbccom2 . (Contributed by Giovanni Mascellani, 31-May-2019)

Ref Expression
Hypothesis sbccom2lem.1 AV
Assertion sbccom2lem [˙A/x]˙[˙B/y]˙φ[˙A/xB/y]˙[˙A/x]˙φ

Proof

Step Hyp Ref Expression
1 sbccom2lem.1 AV
2 sbcan [˙A/x]˙y=Bφ[˙A/x]˙y=B[˙A/x]˙φ
3 sbc5 [˙A/x]˙y=Bφxx=Ay=Bφ
4 1 csbconstgi A/xy=y
5 eqid A/xB=A/xB
6 1 4 5 sbceqi [˙A/x]˙y=By=A/xB
7 6 anbi1i [˙A/x]˙y=B[˙A/x]˙φy=A/xB[˙A/x]˙φ
8 2 3 7 3bitr3i xx=Ay=Bφy=A/xB[˙A/x]˙φ
9 8 exbii yxx=Ay=Bφyy=A/xB[˙A/x]˙φ
10 sbc5 [˙B/y]˙φyy=Bφ
11 10 sbcbii [˙A/x]˙[˙B/y]˙φ[˙A/x]˙yy=Bφ
12 sbc5 [˙A/x]˙yy=Bφxx=Ayy=Bφ
13 11 12 bitri [˙A/x]˙[˙B/y]˙φxx=Ayy=Bφ
14 19.42v yx=Ay=Bφx=Ayy=Bφ
15 14 bicomi x=Ayy=Bφyx=Ay=Bφ
16 15 exbii xx=Ayy=Bφxyx=Ay=Bφ
17 excom xyx=Ay=Bφyxx=Ay=Bφ
18 16 17 bitri xx=Ayy=Bφyxx=Ay=Bφ
19 13 18 bitri [˙A/x]˙[˙B/y]˙φyxx=Ay=Bφ
20 sbc5 [˙A/xB/y]˙[˙A/x]˙φyy=A/xB[˙A/x]˙φ
21 9 19 20 3bitr4i [˙A/x]˙[˙B/y]˙φ[˙A/xB/y]˙[˙A/x]˙φ