Theorem sbcbr1g

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

	Ref	Expression
Assertion	sbcbr1g	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 𝑅 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝑅 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 ⦌ 𝐶 = 𝐶 )
3	2	breq2d	⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝑅 ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝑅 𝐶 ) )
4	1 3	bitrd	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 𝑅 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝑅 𝐶 ) )