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Metamath Proof Explorer

Theorem csbeq1d

Description: Equality deduction for proper substitution into a class. (Contributed by NM, 3-Dec-2005)

Ref Expression
Hypothesis csbeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion csbeq1d ( 𝜑 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = ⦋ 𝐵 / 𝑥 ⦌ 𝐶 )

Proof

Step Hyp Ref Expression
1 csbeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 csbeq1 ⊢ ( 𝐴 = 𝐵 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = ⦋ 𝐵 / 𝑥 ⦌ 𝐶 )
3 1 2 syl ⊢ ( 𝜑 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = ⦋ 𝐵 / 𝑥 ⦌ 𝐶 )