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