Metamath Proof Explorer


Theorem funop

Description: An ordered pair is a function iff it is a singleton of an ordered pair. (Contributed by AV, 20-Sep-2020) A function is a class of ordered pairs, so the fact that an ordered pair may sometimes be itself a function is an "accident" depending on the specific encoding of ordered pairs as classes (in set.mm, the Kuratowski encoding). A more meaningful statement is funsng , as relsnopg is to relop . (New usage is discouraged.)

Ref Expression
Hypotheses funopsn.x ⊢ X ∈ V
funopsn.y ⊢ Y ∈ V
Assertion funop ⊢ Fun ⁡ X Y ↔ ∃ a X = a ∧ X Y = a a

Proof

Step Hyp Ref Expression
1 funopsn.x ⊢ X ∈ V
2 funopsn.y ⊢ Y ∈ V
3 eqid ⊢ X Y = X Y
4 1 2 funopsn ⊢ Fun ⁡ X Y ∧ X Y = X Y → ∃ a X = a ∧ X Y = a a
5 3 4 mpan2 ⊢ Fun ⁡ X Y → ∃ a X = a ∧ X Y = a a
6 vex ⊢ a ∈ V
7 6 6 funsn ⊢ Fun ⁡ a a
8 funeq ⊢ X Y = a a → Fun ⁡ X Y ↔ Fun ⁡ a a
9 7 8 mpbiri ⊢ X Y = a a → Fun ⁡ X Y
10 9 adantl ⊢ X = a ∧ X Y = a a → Fun ⁡ X Y
11 10 exlimiv ⊢ ∃ a X = a ∧ X Y = a a → Fun ⁡ X Y
12 5 11 impbii ⊢ Fun ⁡ X Y ↔ ∃ a X = a ∧ X Y = a a