Metamath Proof Explorer


Theorem bj-rabeqd

Description: Deduction form of rabeq . Note that contrary to rabeq it has no disjoint variable condition. (Contributed by BJ, 27-Apr-2019)

Ref Expression
Hypotheses bj-rabeqd.nf x φ
bj-rabeqd.1 φ A = B
Assertion bj-rabeqd φ x A | ψ = x B | ψ

Proof

Step Hyp Ref Expression
1 bj-rabeqd.nf x φ
2 bj-rabeqd.1 φ A = B
3 eleq2 A = B x A x B
4 3 anbi1d A = B x A ψ x B ψ
5 2 4 syl φ x A ψ x B ψ
6 1 5 bj-rabbida2 φ x A | ψ = x B | ψ