Metamath Proof Explorer


Theorem opfv

Description: Value of a function producing ordered pairs. (Contributed by Thierry Arnoux, 3-Jan-2017)

Ref Expression
Assertion opfv ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → F ⁡ x = 1 st ∘ F ⁡ x 2 nd ∘ F ⁡ x

Proof

Step Hyp Ref Expression
1 simplr ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → ran ⁡ F ⊆ V × V
2 fvelrn ⊢ Fun ⁡ F ∧ x ∈ dom ⁡ F → F ⁡ x ∈ ran ⁡ F
3 2 adantlr ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → F ⁡ x ∈ ran ⁡ F
4 1 3 sseldd ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → F ⁡ x ∈ V × V
5 1st2ndb ⊢ F ⁡ x ∈ V × V ↔ F ⁡ x = 1 st ⁡ F ⁡ x 2 nd ⁡ F ⁡ x
6 4 5 sylib ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → F ⁡ x = 1 st ⁡ F ⁡ x 2 nd ⁡ F ⁡ x
7 fvco ⊢ Fun ⁡ F ∧ x ∈ dom ⁡ F → 1 st ∘ F ⁡ x = 1 st ⁡ F ⁡ x
8 fvco ⊢ Fun ⁡ F ∧ x ∈ dom ⁡ F → 2 nd ∘ F ⁡ x = 2 nd ⁡ F ⁡ x
9 7 8 opeq12d ⊢ Fun ⁡ F ∧ x ∈ dom ⁡ F → 1 st ∘ F ⁡ x 2 nd ∘ F ⁡ x = 1 st ⁡ F ⁡ x 2 nd ⁡ F ⁡ x
10 9 adantlr ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → 1 st ∘ F ⁡ x 2 nd ∘ F ⁡ x = 1 st ⁡ F ⁡ x 2 nd ⁡ F ⁡ x
11 6 10 eqtr4d ⊢ Fun ⁡ F ∧ ran ⁡ F ⊆ V × V ∧ x ∈ dom ⁡ F → F ⁡ x = 1 st ∘ F ⁡ x 2 nd ∘ F ⁡ x