Metamath Proof Explorer


Theorem cdleme18b

Description: Part of proof of Lemma E in Crawley p. 114, 2nd sentence of 4th paragraph. F , G represent f(s), f_s(q) respectively. We show -. f_s(q) =/= q. (Contributed by NM, 12-Oct-2012)

Ref Expression
Hypotheses cdleme18.l ⊢ ≤ ˙ = ≤ K
cdleme18.j ⊢ ∨ ˙ = join ⁡ K
cdleme18.m ⊢ ∧ ˙ = meet ⁡ K
cdleme18.a ⊢ A = Atoms ⁡ K
cdleme18.h ⊢ H = LHyp ⁡ K
cdleme18.u ⊢ U = P ∨ ˙ Q ∧ ˙ W
cdleme18.f ⊢ F = S ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ S ∧ ˙ W
cdleme18.g ⊢ G = P ∨ ˙ Q ∧ ˙ F ∨ ˙ Q ∨ ˙ S ∧ ˙ W
Assertion cdleme18b ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → G ≠ Q

Proof

Step Hyp Ref Expression
1 cdleme18.l ⊢ ≤ ˙ = ≤ K
2 cdleme18.j ⊢ ∨ ˙ = join ⁡ K
3 cdleme18.m ⊢ ∧ ˙ = meet ⁡ K
4 cdleme18.a ⊢ A = Atoms ⁡ K
5 cdleme18.h ⊢ H = LHyp ⁡ K
6 cdleme18.u ⊢ U = P ∨ ˙ Q ∧ ˙ W
7 cdleme18.f ⊢ F = S ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ S ∧ ˙ W
8 cdleme18.g ⊢ G = P ∨ ˙ Q ∧ ˙ F ∨ ˙ Q ∨ ˙ S ∧ ˙ W
9 eqid ⊢ Q = Q
10 oveq2 ⊢ G = Q → Q ∨ ˙ G = Q ∨ ˙ Q
11 simp1l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → K ∈ HL
12 simp22l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ∈ A
13 2 4 hlatjidm ⊢ K ∈ HL ∧ Q ∈ A → Q ∨ ˙ Q = Q
14 11 12 13 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ∨ ˙ Q = Q
15 10 14 sylan9eqr ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ G = Q → Q ∨ ˙ G = Q
16 simp1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → K ∈ HL ∧ W ∈ H
17 simp21l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → P ∈ A
18 simp22 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ∈ A ∧ ¬ Q ≤ ˙ W
19 simp23 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → S ∈ A ∧ ¬ S ≤ ˙ W
20 1 2 4 hlatlej2 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A → Q ≤ ˙ P ∨ ˙ Q
21 11 17 12 20 syl3anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ≤ ˙ P ∨ ˙ Q
22 1 2 3 4 5 6 7 8 cdleme5 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ Q ≤ ˙ P ∨ ˙ Q → Q ∨ ˙ G = P ∨ ˙ Q
23 16 17 12 18 19 21 22 syl132anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ∨ ˙ G = P ∨ ˙ Q
24 23 adantr ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ G = Q → Q ∨ ˙ G = P ∨ ˙ Q
25 15 24 eqtr3d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ G = Q → Q = P ∨ ˙ Q
26 simp3l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → P ≠ Q
27 2 4 2atneat ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → ¬ P ∨ ˙ Q ∈ A
28 11 17 12 26 27 syl13anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → ¬ P ∨ ˙ Q ∈ A
29 nelne2 ⊢ Q ∈ A ∧ ¬ P ∨ ˙ Q ∈ A → Q ≠ P ∨ ˙ Q
30 29 necomd ⊢ Q ∈ A ∧ ¬ P ∨ ˙ Q ∈ A → P ∨ ˙ Q ≠ Q
31 12 28 30 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → P ∨ ˙ Q ≠ Q
32 31 adantr ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ G = Q → P ∨ ˙ Q ≠ Q
33 25 32 eqnetrd ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ G = Q → Q ≠ Q
34 33 ex ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → G = Q → Q ≠ Q
35 34 necon2d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q = Q → G ≠ Q
36 9 35 mpi ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → G ≠ Q