Step |
Hyp |
Ref |
Expression |
1 |
|
atcvr0eq.j |
⊢ ∨ = ( join ‘ 𝐾 ) |
2 |
|
atcvr0eq.z |
⊢ 0 = ( 0. ‘ 𝐾 ) |
3 |
|
atcvr0eq.c |
⊢ 𝐶 = ( ⋖ ‘ 𝐾 ) |
4 |
|
atcvr0eq.a |
⊢ 𝐴 = ( Atoms ‘ 𝐾 ) |
5 |
1 3 4
|
atcvr1 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ≠ 𝑄 ↔ 𝑃 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) |
6 |
2 3 4
|
atcvr0 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ) → 0 𝐶 𝑃 ) |
7 |
6
|
3adant3 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → 0 𝐶 𝑃 ) |
8 |
7
|
biantrurd |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 𝐶 ( 𝑃 ∨ 𝑄 ) ↔ ( 0 𝐶 𝑃 ∧ 𝑃 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) ) |
9 |
5 8
|
bitrd |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ≠ 𝑄 ↔ ( 0 𝐶 𝑃 ∧ 𝑃 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) ) |
10 |
|
simp1 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → 𝐾 ∈ HL ) |
11 |
|
hlop |
⊢ ( 𝐾 ∈ HL → 𝐾 ∈ OP ) |
12 |
11
|
3ad2ant1 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → 𝐾 ∈ OP ) |
13 |
|
eqid |
⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 ) |
14 |
13 2
|
op0cl |
⊢ ( 𝐾 ∈ OP → 0 ∈ ( Base ‘ 𝐾 ) ) |
15 |
12 14
|
syl |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → 0 ∈ ( Base ‘ 𝐾 ) ) |
16 |
13 4
|
atbase |
⊢ ( 𝑃 ∈ 𝐴 → 𝑃 ∈ ( Base ‘ 𝐾 ) ) |
17 |
16
|
3ad2ant2 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → 𝑃 ∈ ( Base ‘ 𝐾 ) ) |
18 |
13 1 4
|
hlatjcl |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ∨ 𝑄 ) ∈ ( Base ‘ 𝐾 ) ) |
19 |
13 3
|
cvrntr |
⊢ ( ( 𝐾 ∈ HL ∧ ( 0 ∈ ( Base ‘ 𝐾 ) ∧ 𝑃 ∈ ( Base ‘ 𝐾 ) ∧ ( 𝑃 ∨ 𝑄 ) ∈ ( Base ‘ 𝐾 ) ) ) → ( ( 0 𝐶 𝑃 ∧ 𝑃 𝐶 ( 𝑃 ∨ 𝑄 ) ) → ¬ 0 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) |
20 |
10 15 17 18 19
|
syl13anc |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( ( 0 𝐶 𝑃 ∧ 𝑃 𝐶 ( 𝑃 ∨ 𝑄 ) ) → ¬ 0 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) |
21 |
9 20
|
sylbid |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ≠ 𝑄 → ¬ 0 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) |
22 |
21
|
necon4ad |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 0 𝐶 ( 𝑃 ∨ 𝑄 ) → 𝑃 = 𝑄 ) ) |
23 |
1 4
|
hlatjidm |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ) → ( 𝑃 ∨ 𝑃 ) = 𝑃 ) |
24 |
23
|
3adant3 |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ∨ 𝑃 ) = 𝑃 ) |
25 |
7 24
|
breqtrrd |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → 0 𝐶 ( 𝑃 ∨ 𝑃 ) ) |
26 |
|
oveq2 |
⊢ ( 𝑃 = 𝑄 → ( 𝑃 ∨ 𝑃 ) = ( 𝑃 ∨ 𝑄 ) ) |
27 |
26
|
breq2d |
⊢ ( 𝑃 = 𝑄 → ( 0 𝐶 ( 𝑃 ∨ 𝑃 ) ↔ 0 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) |
28 |
25 27
|
syl5ibcom |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 = 𝑄 → 0 𝐶 ( 𝑃 ∨ 𝑄 ) ) ) |
29 |
22 28
|
impbid |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) → ( 0 𝐶 ( 𝑃 ∨ 𝑄 ) ↔ 𝑃 = 𝑄 ) ) |