Metamath Proof Explorer


Theorem hashuni

Description: The cardinality of a disjoint union. (Contributed by Mario Carneiro, 24-Jan-2015)

Ref Expression
Hypotheses hashuni.1 ( 𝜑𝐴 ∈ Fin )
hashuni.2 ( 𝜑𝐴 ⊆ Fin )
hashuni.3 ( 𝜑Disj 𝑥𝐴 𝑥 )
Assertion hashuni ( 𝜑 → ( ♯ ‘ 𝐴 ) = Σ 𝑥𝐴 ( ♯ ‘ 𝑥 ) )

Proof

Step Hyp Ref Expression
1 hashuni.1 ( 𝜑𝐴 ∈ Fin )
2 hashuni.2 ( 𝜑𝐴 ⊆ Fin )
3 hashuni.3 ( 𝜑Disj 𝑥𝐴 𝑥 )
4 uniiun 𝐴 = 𝑥𝐴 𝑥
5 4 fveq2i ( ♯ ‘ 𝐴 ) = ( ♯ ‘ 𝑥𝐴 𝑥 )
6 2 sselda ( ( 𝜑𝑥𝐴 ) → 𝑥 ∈ Fin )
7 1 6 3 hashiun ( 𝜑 → ( ♯ ‘ 𝑥𝐴 𝑥 ) = Σ 𝑥𝐴 ( ♯ ‘ 𝑥 ) )
8 5 7 eqtrid ( 𝜑 → ( ♯ ‘ 𝐴 ) = Σ 𝑥𝐴 ( ♯ ‘ 𝑥 ) )