Metamath Proof Explorer


Theorem disjpr2

Description: Two completely distinct unordered pairs are disjoint. (Contributed by Alexander van der Vekens, 11-Nov-2017) (Proof shortened by JJ, 23-Jul-2021)

Ref Expression
Assertion disjpr2 ⊢ A ≠ C ∧ B ≠ C ∧ A ≠ D ∧ B ≠ D → A B ∩ C D = ∅

Proof

Step Hyp Ref Expression
1 df-pr ⊢ C D = C ∪ D
2 1 ineq2i ⊢ A B ∩ C D = A B ∩ C ∪ D
3 indi ⊢ A B ∩ C ∪ D = A B ∩ C ∪ A B ∩ D
4 2 3 eqtri ⊢ A B ∩ C D = A B ∩ C ∪ A B ∩ D
5 df-pr ⊢ A B = A ∪ B
6 5 ineq1i ⊢ A B ∩ C = A ∪ B ∩ C
7 indir ⊢ A ∪ B ∩ C = A ∩ C ∪ B ∩ C
8 6 7 eqtri ⊢ A B ∩ C = A ∩ C ∪ B ∩ C
9 disjsn2 ⊢ A ≠ C → A ∩ C = ∅
10 disjsn2 ⊢ B ≠ C → B ∩ C = ∅
11 9 10 anim12i ⊢ A ≠ C ∧ B ≠ C → A ∩ C = ∅ ∧ B ∩ C = ∅
12 un00 ⊢ A ∩ C = ∅ ∧ B ∩ C = ∅ ↔ A ∩ C ∪ B ∩ C = ∅
13 11 12 sylib ⊢ A ≠ C ∧ B ≠ C → A ∩ C ∪ B ∩ C = ∅
14 8 13 eqtrid ⊢ A ≠ C ∧ B ≠ C → A B ∩ C = ∅
15 14 adantr ⊢ A ≠ C ∧ B ≠ C ∧ A ≠ D ∧ B ≠ D → A B ∩ C = ∅
16 5 ineq1i ⊢ A B ∩ D = A ∪ B ∩ D
17 indir ⊢ A ∪ B ∩ D = A ∩ D ∪ B ∩ D
18 16 17 eqtri ⊢ A B ∩ D = A ∩ D ∪ B ∩ D
19 disjsn2 ⊢ A ≠ D → A ∩ D = ∅
20 disjsn2 ⊢ B ≠ D → B ∩ D = ∅
21 19 20 anim12i ⊢ A ≠ D ∧ B ≠ D → A ∩ D = ∅ ∧ B ∩ D = ∅
22 un00 ⊢ A ∩ D = ∅ ∧ B ∩ D = ∅ ↔ A ∩ D ∪ B ∩ D = ∅
23 21 22 sylib ⊢ A ≠ D ∧ B ≠ D → A ∩ D ∪ B ∩ D = ∅
24 18 23 eqtrid ⊢ A ≠ D ∧ B ≠ D → A B ∩ D = ∅
25 24 adantl ⊢ A ≠ C ∧ B ≠ C ∧ A ≠ D ∧ B ≠ D → A B ∩ D = ∅
26 15 25 uneq12d ⊢ A ≠ C ∧ B ≠ C ∧ A ≠ D ∧ B ≠ D → A B ∩ C ∪ A B ∩ D = ∅ ∪ ∅
27 un0 ⊢ ∅ ∪ ∅ = ∅
28 26 27 eqtrdi ⊢ A ≠ C ∧ B ≠ C ∧ A ≠ D ∧ B ≠ D → A B ∩ C ∪ A B ∩ D = ∅
29 4 28 eqtrid ⊢ A ≠ C ∧ B ≠ C ∧ A ≠ D ∧ B ≠ D → A B ∩ C D = ∅