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SUPPLEMENTARY MATERIAL (USERS' MATHBOXES)
Mathbox for Norm Megill
Construction of a vector space from a Hilbert lattice
cdleme19f
Metamath Proof Explorer
Description: Part of proof of Lemma E in Crawley p. 113, 5th paragraph on p. 114,
line 3. D , F , N , Y , G , O represent s_2,
f(s), f_s(r), t_2, f(t), f_t(r). We prove that if r <_ s \/
t, then f_t(r) = f_t(r). (Contributed by NM , 14-Nov-2012)
Ref
Expression
Hypotheses
cdleme19.l
⊢ ≤ ˙ = ≤ K
cdleme19.j
⊢ ∨ ˙ = join ⁡ K
cdleme19.m
⊢ ∧ ˙ = meet ⁡ K
cdleme19.a
⊢ A = Atoms ⁡ K
cdleme19.h
⊢ H = LHyp ⁡ K
cdleme19.u
⊢ U = P ∨ ˙ Q ∧ ˙ W
cdleme19.f
⊢ F = S ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ S ∧ ˙ W
cdleme19.g
⊢ G = T ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ T ∧ ˙ W
cdleme19.d
⊢ D = R ∨ ˙ S ∧ ˙ W
cdleme19.y
⊢ Y = R ∨ ˙ T ∧ ˙ W
cdleme19.n
⊢ N = P ∨ ˙ Q ∧ ˙ F ∨ ˙ D
cdleme19.o
⊢ O = P ∨ ˙ Q ∧ ˙ G ∨ ˙ Y
Assertion
cdleme19f
⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ T ∈ A ∧ ¬ T ≤ ˙ W ∧ R ∈ A ∧ P ≠ Q ∧ S ≠ T ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ¬ T ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ S ∨ ˙ T → N = O
Proof
Step
Hyp
Ref
Expression
1
cdleme19.l
⊢ ≤ ˙ = ≤ K
2
cdleme19.j
⊢ ∨ ˙ = join ⁡ K
3
cdleme19.m
⊢ ∧ ˙ = meet ⁡ K
4
cdleme19.a
⊢ A = Atoms ⁡ K
5
cdleme19.h
⊢ H = LHyp ⁡ K
6
cdleme19.u
⊢ U = P ∨ ˙ Q ∧ ˙ W
7
cdleme19.f
⊢ F = S ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ S ∧ ˙ W
8
cdleme19.g
⊢ G = T ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ T ∧ ˙ W
9
cdleme19.d
⊢ D = R ∨ ˙ S ∧ ˙ W
10
cdleme19.y
⊢ Y = R ∨ ˙ T ∧ ˙ W
11
cdleme19.n
⊢ N = P ∨ ˙ Q ∧ ˙ F ∨ ˙ D
12
cdleme19.o
⊢ O = P ∨ ˙ Q ∧ ˙ G ∨ ˙ Y
13
1 2 3 4 5 6 7 8 9 10
cdleme19e
⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ T ∈ A ∧ ¬ T ≤ ˙ W ∧ R ∈ A ∧ P ≠ Q ∧ S ≠ T ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ¬ T ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ S ∨ ˙ T → F ∨ ˙ D = G ∨ ˙ Y
14
13
oveq2d
⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ T ∈ A ∧ ¬ T ≤ ˙ W ∧ R ∈ A ∧ P ≠ Q ∧ S ≠ T ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ¬ T ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ S ∨ ˙ T → P ∨ ˙ Q ∧ ˙ F ∨ ˙ D = P ∨ ˙ Q ∧ ˙ G ∨ ˙ Y
15
14 11 12
3eqtr4g
⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ T ∈ A ∧ ¬ T ≤ ˙ W ∧ R ∈ A ∧ P ≠ Q ∧ S ≠ T ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ¬ T ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ P ∨ ˙ Q ∧ R ≤ ˙ S ∨ ˙ T → N = O