Metamath Proof Explorer


Theorem sbcbr2g

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

Ref Expression
Assertion sbcbr2g ⊢ A ∈ V → [˙A / x]˙ B R C ↔ B R ⦋ A / x⦌ C

Proof

Step Hyp Ref Expression
1 sbcbr12g ⊢ A ∈ V → [˙A / x]˙ B R C ↔ ⦋ A / x⦌ B R ⦋ A / x⦌ C
2 csbconstg ⊢ A ∈ V → ⦋ A / x⦌ B = B
3 2 breq1d ⊢ A ∈ V → ⦋ A / x⦌ B R ⦋ A / x⦌ C ↔ B R ⦋ A / x⦌ C
4 1 3 bitrd ⊢ A ∈ V → [˙A / x]˙ B R C ↔ B R ⦋ A / x⦌ C