Metamath Proof Explorer


Theorem funssxp

Description: Two ways of specifying a partial function from A to B . (Contributed by NM, 13-Nov-2007)

Ref Expression
Assertion funssxp ⊢ Fun ⁡ F ∧ F ⊆ A × B ↔ F : dom ⁡ F ⟶ B ∧ dom ⁡ F ⊆ A

Proof

Step Hyp Ref Expression
1 funfn ⊢ Fun ⁡ F ↔ F Fn dom ⁡ F
2 1 biimpi ⊢ Fun ⁡ F → F Fn dom ⁡ F
3 rnss ⊢ F ⊆ A × B → ran ⁡ F ⊆ ran ⁡ A × B
4 rnxpss ⊢ ran ⁡ A × B ⊆ B
5 3 4 sstrdi ⊢ F ⊆ A × B → ran ⁡ F ⊆ B
6 2 5 anim12i ⊢ Fun ⁡ F ∧ F ⊆ A × B → F Fn dom ⁡ F ∧ ran ⁡ F ⊆ B
7 df-f ⊢ F : dom ⁡ F ⟶ B ↔ F Fn dom ⁡ F ∧ ran ⁡ F ⊆ B
8 6 7 sylibr ⊢ Fun ⁡ F ∧ F ⊆ A × B → F : dom ⁡ F ⟶ B
9 dmss ⊢ F ⊆ A × B → dom ⁡ F ⊆ dom ⁡ A × B
10 dmxpss ⊢ dom ⁡ A × B ⊆ A
11 9 10 sstrdi ⊢ F ⊆ A × B → dom ⁡ F ⊆ A
12 11 adantl ⊢ Fun ⁡ F ∧ F ⊆ A × B → dom ⁡ F ⊆ A
13 8 12 jca ⊢ Fun ⁡ F ∧ F ⊆ A × B → F : dom ⁡ F ⟶ B ∧ dom ⁡ F ⊆ A
14 ffun ⊢ F : dom ⁡ F ⟶ B → Fun ⁡ F
15 14 adantr ⊢ F : dom ⁡ F ⟶ B ∧ dom ⁡ F ⊆ A → Fun ⁡ F
16 fssxp ⊢ F : dom ⁡ F ⟶ B → F ⊆ dom ⁡ F × B
17 xpss1 ⊢ dom ⁡ F ⊆ A → dom ⁡ F × B ⊆ A × B
18 16 17 sylan9ss ⊢ F : dom ⁡ F ⟶ B ∧ dom ⁡ F ⊆ A → F ⊆ A × B
19 15 18 jca ⊢ F : dom ⁡ F ⟶ B ∧ dom ⁡ F ⊆ A → Fun ⁡ F ∧ F ⊆ A × B
20 13 19 impbii ⊢ Fun ⁡ F ∧ F ⊆ A × B ↔ F : dom ⁡ F ⟶ B ∧ dom ⁡ F ⊆ A