Metamath Proof Explorer


Theorem funcnvpr

Description: The converse pair of ordered pairs is a function if the second members are different. Note that the second members need not be sets. (Contributed by AV, 23-Jan-2021)

Ref Expression
Assertion funcnvpr ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → Fun ⁡ A B C D -1

Proof

Step Hyp Ref Expression
1 funcnvsn ⊢ Fun ⁡ A B -1
2 funcnvsn ⊢ Fun ⁡ C D -1
3 1 2 pm3.2i ⊢ Fun ⁡ A B -1 ∧ Fun ⁡ C D -1
4 df-rn ⊢ ran ⁡ A B = dom ⁡ A B -1
5 rnsnopg ⊢ A ∈ U → ran ⁡ A B = B
6 4 5 eqtr3id ⊢ A ∈ U → dom ⁡ A B -1 = B
7 df-rn ⊢ ran ⁡ C D = dom ⁡ C D -1
8 rnsnopg ⊢ C ∈ V → ran ⁡ C D = D
9 7 8 eqtr3id ⊢ C ∈ V → dom ⁡ C D -1 = D
10 6 9 ineqan12d ⊢ A ∈ U ∧ C ∈ V → dom ⁡ A B -1 ∩ dom ⁡ C D -1 = B ∩ D
11 10 3adant3 ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → dom ⁡ A B -1 ∩ dom ⁡ C D -1 = B ∩ D
12 disjsn2 ⊢ B ≠ D → B ∩ D = ∅
13 12 3ad2ant3 ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → B ∩ D = ∅
14 11 13 eqtrd ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → dom ⁡ A B -1 ∩ dom ⁡ C D -1 = ∅
15 funun ⊢ Fun ⁡ A B -1 ∧ Fun ⁡ C D -1 ∧ dom ⁡ A B -1 ∩ dom ⁡ C D -1 = ∅ → Fun ⁡ A B -1 ∪ C D -1
16 3 14 15 sylancr ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → Fun ⁡ A B -1 ∪ C D -1
17 df-pr ⊢ A B C D = A B ∪ C D
18 17 cnveqi ⊢ A B C D -1 = A B ∪ C D -1
19 cnvun ⊢ A B ∪ C D -1 = A B -1 ∪ C D -1
20 18 19 eqtri ⊢ A B C D -1 = A B -1 ∪ C D -1
21 20 funeqi ⊢ Fun ⁡ A B C D -1 ↔ Fun ⁡ A B -1 ∪ C D -1
22 16 21 sylibr ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → Fun ⁡ A B C D -1