Metamath Proof Explorer


Theorem hashuni

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

Ref Expression
Hypotheses hashuni.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
hashuni.2 ⊢ ( 𝜑 → 𝐴 ⊆ Fin )
hashuni.3 ⊢ ( 𝜑 → Disj 𝑥 ∈ 𝐴 𝑥 )
Assertion hashuni ( 𝜑 → ( ♯ ‘ ∪ 𝐴 ) = Σ 𝑥 ∈ 𝐴 ( ♯ ‘ 𝑥 ) )

Proof

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