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Metamath Proof Explorer

Theorem ensymd

Description: Symmetry of equinumerosity. Deduction form of ensym . (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypothesis ensymd.1 ⊢ ( 𝜑 → 𝐴 ≈ 𝐵 )
Assertion ensymd ( 𝜑 → 𝐵 ≈ 𝐴 )

Proof

Step Hyp Ref Expression
1 ensymd.1 ⊢ ( 𝜑 → 𝐴 ≈ 𝐵 )
2 ensym ⊢ ( 𝐴 ≈ 𝐵 → 𝐵 ≈ 𝐴 )
3 1 2 syl ⊢ ( 𝜑 → 𝐵 ≈ 𝐴 )