Metamath Proof Explorer


Theorem hlatexch4

Description: Exchange 2 atoms. (Contributed by NM, 13-May-2013)

Ref Expression
Hypotheses hlatexch4.j ⊢ ∨ ˙ = join ⁡ K
hlatexch4.a ⊢ A = Atoms ⁡ K
Assertion hlatexch4 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∨ ˙ R = Q ∨ ˙ S

Proof

Step Hyp Ref Expression
1 hlatexch4.j ⊢ ∨ ˙ = join ⁡ K
2 hlatexch4.a ⊢ A = Atoms ⁡ K
3 simp11 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → K ∈ HL
4 simp2l ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → R ∈ A
5 simp2r ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → S ∈ A
6 eqid ⊢ ≤ K = ≤ K
7 6 1 2 hlatlej2 ⊢ K ∈ HL ∧ R ∈ A ∧ S ∈ A → S ≤ K R ∨ ˙ S
8 3 4 5 7 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → S ≤ K R ∨ ˙ S
9 simp33 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∨ ˙ Q = R ∨ ˙ S
10 8 9 breqtrrd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → S ≤ K P ∨ ˙ Q
11 simp12 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∈ A
12 simp13 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → Q ∈ A
13 simp32 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → Q ≠ S
14 13 necomd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → S ≠ Q
15 6 1 2 hlatexch2 ⊢ K ∈ HL ∧ S ∈ A ∧ P ∈ A ∧ Q ∈ A ∧ S ≠ Q → S ≤ K P ∨ ˙ Q → P ≤ K S ∨ ˙ Q
16 3 5 11 12 14 15 syl131anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → S ≤ K P ∨ ˙ Q → P ≤ K S ∨ ˙ Q
17 10 16 mpd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ≤ K S ∨ ˙ Q
18 1 2 hlatjcom ⊢ K ∈ HL ∧ S ∈ A ∧ Q ∈ A → S ∨ ˙ Q = Q ∨ ˙ S
19 3 5 12 18 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → S ∨ ˙ Q = Q ∨ ˙ S
20 17 19 breqtrd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ≤ K Q ∨ ˙ S
21 6 1 2 hlatlej2 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A → Q ≤ K P ∨ ˙ Q
22 3 11 12 21 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → Q ≤ K P ∨ ˙ Q
23 22 9 breqtrd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → Q ≤ K R ∨ ˙ S
24 6 1 2 hlatexch2 ⊢ K ∈ HL ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ Q ≠ S → Q ≤ K R ∨ ˙ S → R ≤ K Q ∨ ˙ S
25 3 12 4 5 13 24 syl131anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → Q ≤ K R ∨ ˙ S → R ≤ K Q ∨ ˙ S
26 23 25 mpd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → R ≤ K Q ∨ ˙ S
27 3 hllatd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → K ∈ Lat
28 eqid ⊢ Base K = Base K
29 28 2 atbase ⊢ P ∈ A → P ∈ Base K
30 11 29 syl ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∈ Base K
31 28 2 atbase ⊢ R ∈ A → R ∈ Base K
32 4 31 syl ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → R ∈ Base K
33 28 1 2 hlatjcl ⊢ K ∈ HL ∧ Q ∈ A ∧ S ∈ A → Q ∨ ˙ S ∈ Base K
34 3 12 5 33 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → Q ∨ ˙ S ∈ Base K
35 28 6 1 latjle12 ⊢ K ∈ Lat ∧ P ∈ Base K ∧ R ∈ Base K ∧ Q ∨ ˙ S ∈ Base K → P ≤ K Q ∨ ˙ S ∧ R ≤ K Q ∨ ˙ S ↔ P ∨ ˙ R ≤ K Q ∨ ˙ S
36 27 30 32 34 35 syl13anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ≤ K Q ∨ ˙ S ∧ R ≤ K Q ∨ ˙ S ↔ P ∨ ˙ R ≤ K Q ∨ ˙ S
37 20 26 36 mpbi2and ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∨ ˙ R ≤ K Q ∨ ˙ S
38 simp31 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ≠ R
39 6 1 2 ps-1 ⊢ K ∈ HL ∧ P ∈ A ∧ R ∈ A ∧ P ≠ R ∧ Q ∈ A ∧ S ∈ A → P ∨ ˙ R ≤ K Q ∨ ˙ S ↔ P ∨ ˙ R = Q ∨ ˙ S
40 3 11 4 38 12 5 39 syl132anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∨ ˙ R ≤ K Q ∨ ˙ S ↔ P ∨ ˙ R = Q ∨ ˙ S
41 37 40 mpbid ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ S ∈ A ∧ P ≠ R ∧ Q ≠ S ∧ P ∨ ˙ Q = R ∨ ˙ S → P ∨ ˙ R = Q ∨ ˙ S