Metamath Proof Explorer


Theorem sbcor

Description: Distribution of class substitution over disjunction. (Contributed by NM, 31-Dec-2016) (Revised by NM, 17-Aug-2018)

Ref Expression
Assertion sbcor [˙A/x]˙φψ[˙A/x]˙φ[˙A/x]˙ψ

Proof

Step Hyp Ref Expression
1 sbcex [˙A/x]˙φψAV
2 sbcex [˙A/x]˙φAV
3 sbcex [˙A/x]˙ψAV
4 2 3 jaoi [˙A/x]˙φ[˙A/x]˙ψAV
5 dfsbcq2 y=Ayxφψ[˙A/x]˙φψ
6 dfsbcq2 y=Ayxφ[˙A/x]˙φ
7 dfsbcq2 y=Ayxψ[˙A/x]˙ψ
8 6 7 orbi12d y=Ayxφyxψ[˙A/x]˙φ[˙A/x]˙ψ
9 sbor yxφψyxφyxψ
10 5 8 9 vtoclbg AV[˙A/x]˙φψ[˙A/x]˙φ[˙A/x]˙ψ
11 1 4 10 pm5.21nii [˙A/x]˙φψ[˙A/x]˙φ[˙A/x]˙ψ