Metamath Proof Explorer


Theorem sbcbid

Description: Formula-building deduction for class substitution. (Contributed by NM, 29-Dec-2014)

Ref Expression
Hypotheses sbcbid.1 xφ
sbcbid.2 φψχ
Assertion sbcbid φ[˙A/x]˙ψ[˙A/x]˙χ

Proof

Step Hyp Ref Expression
1 sbcbid.1 xφ
2 sbcbid.2 φψχ
3 1 2 abbid φx|ψ=x|χ
4 3 eleq2d φAx|ψAx|χ
5 df-sbc [˙A/x]˙ψAx|ψ
6 df-sbc [˙A/x]˙χAx|χ
7 4 5 6 3bitr4g φ[˙A/x]˙ψ[˙A/x]˙χ