Metamath Proof Explorer


Theorem fex2

Description: A function with bounded domain and codomain is a set. This version of fex is proven without the Axiom of Replacement ax-rep , but depends on ax-un , which is not required for the proof of fex . (Contributed by Mario Carneiro, 24-Jun-2015)

Ref Expression
Assertion fex2 ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → 𝐹 ∈ V )

Proof

Step Hyp Ref Expression
1 xpexg ⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( 𝐴 × 𝐵 ) ∈ V )
2 1 3adant1 ⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( 𝐴 × 𝐵 ) ∈ V )
3 fssxp ⊢ ( 𝐹 : 𝐴 ⟶ 𝐵 → 𝐹 ⊆ ( 𝐴 × 𝐵 ) )
4 3 3ad2ant1 ⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → 𝐹 ⊆ ( 𝐴 × 𝐵 ) )
5 2 4 ssexd ⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → 𝐹 ∈ V )