Theorem sbceq1g

Description: Move proper substitution to first argument of an equality. (Contributed by NM, 30-Nov-2005)

	Ref	Expression
Assertion	sbceq1g	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		sbceqg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) )
2		csbconstg	⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = 𝐶 )
3	2	eqeq2d	⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐶 ) )
4	1 3	bitrd	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐶 ) )