Metamath Proof Explorer


Theorem equcomd

Description: Deduction form of equcom , symmetry of equality. For the versions for classes, see eqcom and eqcomd . (Contributed by BJ, 6-Oct-2019)

Ref Expression
Hypothesis equcomd.1 ⊢ ( 𝜑 → 𝑥 = 𝑦 )
Assertion equcomd ( 𝜑 → 𝑦 = 𝑥 )

Proof

Step Hyp Ref Expression
1 equcomd.1 ⊢ ( 𝜑 → 𝑥 = 𝑦 )
2 equcom ⊢ ( 𝑥 = 𝑦 ↔ 𝑦 = 𝑥 )
3 1 2 sylib ⊢ ( 𝜑 → 𝑦 = 𝑥 )