Metamath Proof Explorer


Theorem hashfn

Description: A function is equinumerous to its domain. (Contributed by Mario Carneiro, 12-Mar-2015)

Ref Expression
Assertion hashfn ( 𝐹 Fn 𝐴 → ( ♯ ‘ 𝐹 ) = ( ♯ ‘ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 fndmeng ( ( 𝐹 Fn 𝐴𝐴 ∈ V ) → 𝐴𝐹 )
2 ensym ( 𝐴𝐹𝐹𝐴 )
3 hasheni ( 𝐹𝐴 → ( ♯ ‘ 𝐹 ) = ( ♯ ‘ 𝐴 ) )
4 1 2 3 3syl ( ( 𝐹 Fn 𝐴𝐴 ∈ V ) → ( ♯ ‘ 𝐹 ) = ( ♯ ‘ 𝐴 ) )
5 dmexg ( 𝐹 ∈ V → dom 𝐹 ∈ V )
6 fndm ( 𝐹 Fn 𝐴 → dom 𝐹 = 𝐴 )
7 6 eleq1d ( 𝐹 Fn 𝐴 → ( dom 𝐹 ∈ V ↔ 𝐴 ∈ V ) )
8 5 7 syl5ib ( 𝐹 Fn 𝐴 → ( 𝐹 ∈ V → 𝐴 ∈ V ) )
9 8 con3dimp ( ( 𝐹 Fn 𝐴 ∧ ¬ 𝐴 ∈ V ) → ¬ 𝐹 ∈ V )
10 fvprc ( ¬ 𝐹 ∈ V → ( ♯ ‘ 𝐹 ) = ∅ )
11 9 10 syl ( ( 𝐹 Fn 𝐴 ∧ ¬ 𝐴 ∈ V ) → ( ♯ ‘ 𝐹 ) = ∅ )
12 fvprc ( ¬ 𝐴 ∈ V → ( ♯ ‘ 𝐴 ) = ∅ )
13 12 adantl ( ( 𝐹 Fn 𝐴 ∧ ¬ 𝐴 ∈ V ) → ( ♯ ‘ 𝐴 ) = ∅ )
14 11 13 eqtr4d ( ( 𝐹 Fn 𝐴 ∧ ¬ 𝐴 ∈ V ) → ( ♯ ‘ 𝐹 ) = ( ♯ ‘ 𝐴 ) )
15 4 14 pm2.61dan ( 𝐹 Fn 𝐴 → ( ♯ ‘ 𝐹 ) = ( ♯ ‘ 𝐴 ) )