Metamath Proof Explorer


Theorem fnxpdmdm

Description: The domain of the domain of a function over a Cartesian square. (Contributed by AV, 13-Jan-2020)

Ref Expression
Assertion fnxpdmdm ⊢ F Fn A × A → dom ⁡ dom ⁡ F = A

Proof

Step Hyp Ref Expression
1 fndm ⊢ F Fn A × A → dom ⁡ F = A × A
2 dmeq ⊢ dom ⁡ F = A × A → dom ⁡ dom ⁡ F = dom ⁡ A × A
3 dmxpid ⊢ dom ⁡ A × A = A
4 2 3 eqtrdi ⊢ dom ⁡ F = A × A → dom ⁡ dom ⁡ F = A
5 1 4 syl ⊢ F Fn A × A → dom ⁡ dom ⁡ F = A