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Metamath Proof Explorer

Theorem sbcbr12g

Description: Move substitution in and out of a binary relation. (Contributed by NM, 13-Dec-2005)

Ref Expression
Assertion sbcbr12g ⊢ A ∈ V → [˙A / x]˙ B R C ↔ ⦋ A / x⦌ B R ⦋ A / x⦌ C

Proof

Step Hyp Ref Expression
1 sbcbr123 ⊢ [˙A / x]˙ B R C ↔ ⦋ A / x⦌ B ⦋ A / x⦌ R ⦋ A / x⦌ C
2 csbconstg ⊢ A ∈ V → ⦋ A / x⦌ R = R
3 2 breqd ⊢ A ∈ V → ⦋ A / x⦌ B ⦋ A / x⦌ R ⦋ A / x⦌ C ↔ ⦋ A / x⦌ B R ⦋ A / x⦌ C
4 1 3 bitrid ⊢ A ∈ V → [˙A / x]˙ B R C ↔ ⦋ A / x⦌ B R ⦋ A / x⦌ C