Metamath Proof Explorer


Theorem cdleml6

Description: Part of proof of Lemma L of Crawley p. 120. TODO: fix comment. (Contributed by NM, 11-Aug-2013)

Ref Expression
Hypotheses cdleml6.b ⊢ B = Base K
cdleml6.j ⊢ ∨ ˙ = join ⁡ K
cdleml6.m ⊢ ∧ ˙ = meet ⁡ K
cdleml6.h ⊢ H = LHyp ⁡ K
cdleml6.t ⊢ T = LTrn ⁡ K ⁡ W
cdleml6.r ⊢ R = trL ⁡ K ⁡ W
cdleml6.p ⊢ Q = oc ⁡ K ⁡ W
cdleml6.z ⊢ Z = Q ∨ ˙ R ⁡ b ∧ ˙ h ⁡ Q ∨ ˙ R ⁡ b ∘ s ⁡ h -1
cdleml6.y ⊢ Y = Q ∨ ˙ R ⁡ g ∧ ˙ Z ∨ ˙ R ⁡ g ∘ b -1
cdleml6.x ⊢ X = ι z ∈ T | ∀ b ∈ T b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ s ⁡ h ∧ R ⁡ b ≠ R ⁡ g → z ⁡ Q = Y
cdleml6.u ⊢ U = g ∈ T ⟼ if s ⁡ h = h g X
cdleml6.e ⊢ E = TEndo ⁡ K ⁡ W
cdleml6.o ⊢ 0 ˙ = f ∈ T ⟼ I ↾ B
Assertion cdleml6 ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → U ∈ E ∧ U ⁡ s ⁡ h = h

Proof

Step Hyp Ref Expression
1 cdleml6.b ⊢ B = Base K
2 cdleml6.j ⊢ ∨ ˙ = join ⁡ K
3 cdleml6.m ⊢ ∧ ˙ = meet ⁡ K
4 cdleml6.h ⊢ H = LHyp ⁡ K
5 cdleml6.t ⊢ T = LTrn ⁡ K ⁡ W
6 cdleml6.r ⊢ R = trL ⁡ K ⁡ W
7 cdleml6.p ⊢ Q = oc ⁡ K ⁡ W
8 cdleml6.z ⊢ Z = Q ∨ ˙ R ⁡ b ∧ ˙ h ⁡ Q ∨ ˙ R ⁡ b ∘ s ⁡ h -1
9 cdleml6.y ⊢ Y = Q ∨ ˙ R ⁡ g ∧ ˙ Z ∨ ˙ R ⁡ g ∘ b -1
10 cdleml6.x ⊢ X = ι z ∈ T | ∀ b ∈ T b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ s ⁡ h ∧ R ⁡ b ≠ R ⁡ g → z ⁡ Q = Y
11 cdleml6.u ⊢ U = g ∈ T ⟼ if s ⁡ h = h g X
12 cdleml6.e ⊢ E = TEndo ⁡ K ⁡ W
13 cdleml6.o ⊢ 0 ˙ = f ∈ T ⟼ I ↾ B
14 simp1 ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → K ∈ HL ∧ W ∈ H
15 simp3l ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → s ∈ E
16 simp2 ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → h ∈ T
17 4 5 12 tendocl ⊢ K ∈ HL ∧ W ∈ H ∧ s ∈ E ∧ h ∈ T → s ⁡ h ∈ T
18 14 15 16 17 syl3anc ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → s ⁡ h ∈ T
19 1 4 5 6 12 13 tendotr ⊢ K ∈ HL ∧ W ∈ H ∧ s ∈ E ∧ s ≠ 0 ˙ ∧ h ∈ T → R ⁡ s ⁡ h = R ⁡ h
20 19 3com23 ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → R ⁡ s ⁡ h = R ⁡ h
21 eqid ⊢ oc ⁡ K = oc ⁡ K
22 eqid ⊢ Atoms ⁡ K = Atoms ⁡ K
23 1 2 3 21 22 4 5 6 7 8 9 10 11 12 cdlemk56w ⊢ K ∈ HL ∧ W ∈ H ∧ s ⁡ h ∈ T ∧ h ∈ T ∧ R ⁡ s ⁡ h = R ⁡ h → U ∈ E ∧ U ⁡ s ⁡ h = h
24 14 18 16 20 23 syl121anc ⊢ K ∈ HL ∧ W ∈ H ∧ h ∈ T ∧ s ∈ E ∧ s ≠ 0 ˙ → U ∈ E ∧ U ⁡ s ⁡ h = h