Metamath Proof Explorer


Theorem sbccomieg

Description: Commute two explicit substitutions, using an implicit substitution to rewrite the exiting substitution. (Contributed by Stefan O'Rear, 11-Oct-2014) (Revised by Mario Carneiro, 11-Dec-2016)

Ref Expression
Hypothesis sbccomieg.1 a=AB=C
Assertion sbccomieg [˙A/a]˙[˙B/b]˙φ[˙C/b]˙[˙A/a]˙φ

Proof

Step Hyp Ref Expression
1 sbccomieg.1 a=AB=C
2 sbcex [˙A/a]˙[˙B/b]˙φAV
3 spesbc [˙C/b]˙[˙A/a]˙φb[˙A/a]˙φ
4 sbcex [˙A/a]˙φAV
5 4 exlimiv b[˙A/a]˙φAV
6 3 5 syl [˙C/b]˙[˙A/a]˙φAV
7 nfcv _aC
8 nfsbc1v a[˙A/a]˙φ
9 7 8 nfsbcw a[˙C/b]˙[˙A/a]˙φ
10 sbceq1a a=Aφ[˙A/a]˙φ
11 1 10 sbceqbid a=A[˙B/b]˙φ[˙C/b]˙[˙A/a]˙φ
12 9 11 sbciegf AV[˙A/a]˙[˙B/b]˙φ[˙C/b]˙[˙A/a]˙φ
13 2 6 12 pm5.21nii [˙A/a]˙[˙B/b]˙φ[˙C/b]˙[˙A/a]˙φ