Metamath Proof Explorer

Theorem cvlatexchb2

Description: A version of cvlexchb2 for atoms. (Contributed by NM, 5-Nov-2012)

Ref Expression
Hypotheses cvlatexch.l ⊢ ≤ = ( le ‘ 𝐾 )
cvlatexch.j ⊢ ∨ = ( join ‘ 𝐾 )
cvlatexch.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
Assertion cvlatexchb2 ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( 𝑃 ≤ ( 𝑄 ∨ 𝑅 ) ↔ ( 𝑃 ∨ 𝑅 ) = ( 𝑄 ∨ 𝑅 ) ) )

Proof

Step Hyp Ref Expression
1 cvlatexch.l ⊢ ≤ = ( le ‘ 𝐾 )
2 cvlatexch.j ⊢ ∨ = ( join ‘ 𝐾 )
3 cvlatexch.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
4 1 2 3 cvlatexchb1 ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( 𝑃 ≤ ( 𝑅 ∨ 𝑄 ) ↔ ( 𝑅 ∨ 𝑃 ) = ( 𝑅 ∨ 𝑄 ) ) )
5 cvllat ⊢ ( 𝐾 ∈ CvLat → 𝐾 ∈ Lat )
6 5 3ad2ant1 ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝐾 ∈ Lat )
7 simp22 ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝑄 ∈ 𝐴 )
8 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
9 8 3 atbase ⊢ ( 𝑄 ∈ 𝐴 → 𝑄 ∈ ( Base ‘ 𝐾 ) )
10 7 9 syl ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝑄 ∈ ( Base ‘ 𝐾 ) )
11 simp23 ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝑅 ∈ 𝐴 )
12 8 3 atbase ⊢ ( 𝑅 ∈ 𝐴 → 𝑅 ∈ ( Base ‘ 𝐾 ) )
13 11 12 syl ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝑅 ∈ ( Base ‘ 𝐾 ) )
14 8 2 latjcom ⊢ ( ( 𝐾 ∈ Lat ∧ 𝑄 ∈ ( Base ‘ 𝐾 ) ∧ 𝑅 ∈ ( Base ‘ 𝐾 ) ) → ( 𝑄 ∨ 𝑅 ) = ( 𝑅 ∨ 𝑄 ) )
15 6 10 13 14 syl3anc ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( 𝑄 ∨ 𝑅 ) = ( 𝑅 ∨ 𝑄 ) )
16 15 breq2d ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( 𝑃 ≤ ( 𝑄 ∨ 𝑅 ) ↔ 𝑃 ≤ ( 𝑅 ∨ 𝑄 ) ) )
17 simp21 ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝑃 ∈ 𝐴 )
18 8 3 atbase ⊢ ( 𝑃 ∈ 𝐴 → 𝑃 ∈ ( Base ‘ 𝐾 ) )
19 17 18 syl ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → 𝑃 ∈ ( Base ‘ 𝐾 ) )
20 8 2 latjcom ⊢ ( ( 𝐾 ∈ Lat ∧ 𝑃 ∈ ( Base ‘ 𝐾 ) ∧ 𝑅 ∈ ( Base ‘ 𝐾 ) ) → ( 𝑃 ∨ 𝑅 ) = ( 𝑅 ∨ 𝑃 ) )
21 6 19 13 20 syl3anc ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( 𝑃 ∨ 𝑅 ) = ( 𝑅 ∨ 𝑃 ) )
22 21 15 eqeq12d ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( ( 𝑃 ∨ 𝑅 ) = ( 𝑄 ∨ 𝑅 ) ↔ ( 𝑅 ∨ 𝑃 ) = ( 𝑅 ∨ 𝑄 ) ) )
23 4 16 22 3bitr4d ⊢ ( ( 𝐾 ∈ CvLat ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ 𝑃 ≠ 𝑅 ) → ( 𝑃 ≤ ( 𝑄 ∨ 𝑅 ) ↔ ( 𝑃 ∨ 𝑅 ) = ( 𝑄 ∨ 𝑅 ) ) )