Metamath Proof Explorer


Theorem funcnvqp

Description: The converse quadruple of ordered pairs is a function if the second members are pairwise different. Note that the second members need not be sets. (Contributed by AV, 23-Jan-2021) (Proof shortened by JJ, 14-Jul-2021)

Ref Expression
Assertion funcnvqp ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → Fun ⁡ A B C D ∪ E F G H -1

Proof

Step Hyp Ref Expression
1 funcnvpr ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → Fun ⁡ A B C D -1
2 1 3expa ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D → Fun ⁡ A B C D -1
3 2 3ad2antr1 ⊢ A ∈ U ∧ C ∈ V ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H → Fun ⁡ A B C D -1
4 3 ad2ant2r ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ F ≠ H → Fun ⁡ A B C D -1
5 4 3adantr2 ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → Fun ⁡ A B C D -1
6 funcnvpr ⊢ E ∈ W ∧ G ∈ T ∧ F ≠ H → Fun ⁡ E F G H -1
7 6 3expa ⊢ E ∈ W ∧ G ∈ T ∧ F ≠ H → Fun ⁡ E F G H -1
8 7 ad2ant2l ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ F ≠ H → Fun ⁡ E F G H -1
9 8 3adantr2 ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → Fun ⁡ E F G H -1
10 df-rn ⊢ ran ⁡ A B C D = dom ⁡ A B C D -1
11 rnpropg ⊢ A ∈ U ∧ C ∈ V → ran ⁡ A B C D = B D
12 10 11 eqtr3id ⊢ A ∈ U ∧ C ∈ V → dom ⁡ A B C D -1 = B D
13 df-rn ⊢ ran ⁡ E F G H = dom ⁡ E F G H -1
14 rnpropg ⊢ E ∈ W ∧ G ∈ T → ran ⁡ E F G H = F H
15 13 14 eqtr3id ⊢ E ∈ W ∧ G ∈ T → dom ⁡ E F G H -1 = F H
16 12 15 ineqan12d ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T → dom ⁡ A B C D -1 ∩ dom ⁡ E F G H -1 = B D ∩ F H
17 disjpr2 ⊢ B ≠ F ∧ D ≠ F ∧ B ≠ H ∧ D ≠ H → B D ∩ F H = ∅
18 17 an4s ⊢ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H → B D ∩ F H = ∅
19 18 3adantl1 ⊢ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H → B D ∩ F H = ∅
20 19 3adant3 ⊢ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → B D ∩ F H = ∅
21 16 20 sylan9eq ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → dom ⁡ A B C D -1 ∩ dom ⁡ E F G H -1 = ∅
22 funun ⊢ Fun ⁡ A B C D -1 ∧ Fun ⁡ E F G H -1 ∧ dom ⁡ A B C D -1 ∩ dom ⁡ E F G H -1 = ∅ → Fun ⁡ A B C D -1 ∪ E F G H -1
23 5 9 21 22 syl21anc ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → Fun ⁡ A B C D -1 ∪ E F G H -1
24 cnvun ⊢ A B C D ∪ E F G H -1 = A B C D -1 ∪ E F G H -1
25 24 funeqi ⊢ Fun ⁡ A B C D ∪ E F G H -1 ↔ Fun ⁡ A B C D -1 ∪ E F G H -1
26 23 25 sylibr ⊢ A ∈ U ∧ C ∈ V ∧ E ∈ W ∧ G ∈ T ∧ B ≠ D ∧ B ≠ F ∧ B ≠ H ∧ D ≠ F ∧ D ≠ H ∧ F ≠ H → Fun ⁡ A B C D ∪ E F G H -1