Metamath Proof Explorer


Theorem sbequi

Description: An equality theorem for substitution. (Contributed by NM, 14-May-1993) (Proof shortened by Wolf Lammen, 15-Sep-2018) (Proof shortened by Steven Nguyen, 7-Jul-2023)

Ref Expression
Assertion sbequi ( 𝑥 = 𝑦 → ( [ 𝑥 / 𝑧 ] 𝜑 → [ 𝑦 / 𝑧 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 sbequ ( 𝑥 = 𝑦 → ( [ 𝑥 / 𝑧 ] 𝜑 ↔ [ 𝑦 / 𝑧 ] 𝜑 ) )
2 1 biimpd ( 𝑥 = 𝑦 → ( [ 𝑥 / 𝑧 ] 𝜑 → [ 𝑦 / 𝑧 ] 𝜑 ) )