Metamath Proof Explorer


Theorem tpossym

Description: Two ways to say a function is symmetric. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Assertion tpossym ⊢ F Fn A × A → tpos F = F ↔ ∀ x ∈ A ∀ y ∈ A x F y = y F x

Proof

Step Hyp Ref Expression
1 tposfn ⊢ F Fn A × A → tpos F Fn A × A
2 eqfnov2 ⊢ tpos F Fn A × A ∧ F Fn A × A → tpos F = F ↔ ∀ x ∈ A ∀ y ∈ A x tpos F y = x F y
3 1 2 mpancom ⊢ F Fn A × A → tpos F = F ↔ ∀ x ∈ A ∀ y ∈ A x tpos F y = x F y
4 eqcom ⊢ x tpos F y = x F y ↔ x F y = x tpos F y
5 ovtpos ⊢ x tpos F y = y F x
6 5 eqeq2i ⊢ x F y = x tpos F y ↔ x F y = y F x
7 4 6 bitri ⊢ x tpos F y = x F y ↔ x F y = y F x
8 7 2ralbii ⊢ ∀ x ∈ A ∀ y ∈ A x tpos F y = x F y ↔ ∀ x ∈ A ∀ y ∈ A x F y = y F x
9 3 8 bitrdi ⊢ F Fn A × A → tpos F = F ↔ ∀ x ∈ A ∀ y ∈ A x F y = y F x