Metamath Proof Explorer


Theorem elpm2g

Description: The predicate "is a partial function." (Contributed by NM, 31-Dec-2013)

Ref Expression
Assertion elpm2g ( ( 𝐴𝑉𝐵𝑊 ) → ( 𝐹 ∈ ( 𝐴pm 𝐵 ) ↔ ( 𝐹 : dom 𝐹𝐴 ∧ dom 𝐹𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 elpmg ( ( 𝐴𝑉𝐵𝑊 ) → ( 𝐹 ∈ ( 𝐴pm 𝐵 ) ↔ ( Fun 𝐹𝐹 ⊆ ( 𝐵 × 𝐴 ) ) ) )
2 funssxp ( ( Fun 𝐹𝐹 ⊆ ( 𝐵 × 𝐴 ) ) ↔ ( 𝐹 : dom 𝐹𝐴 ∧ dom 𝐹𝐵 ) )
3 1 2 syl6bb ( ( 𝐴𝑉𝐵𝑊 ) → ( 𝐹 ∈ ( 𝐴pm 𝐵 ) ↔ ( 𝐹 : dom 𝐹𝐴 ∧ dom 𝐹𝐵 ) ) )