Metamath Proof Explorer


Theorem dihjust

Description: Part of proof after Lemma N of Crawley p. 122 line 4, "the definition of phi(x) is independent of the atom q." (Contributed by NM, 2-Mar-2014)

Ref Expression
Hypotheses dihjust.b ⊢ B = Base K
dihjust.l ⊢ ≤ ˙ = ≤ K
dihjust.j ⊢ ∨ ˙ = join ⁡ K
dihjust.m ⊢ ∧ ˙ = meet ⁡ K
dihjust.a ⊢ A = Atoms ⁡ K
dihjust.h ⊢ H = LHyp ⁡ K
dihjust.i ⊢ I = DIsoB ⁡ K ⁡ W
dihjust.J ⊢ J = DIsoC ⁡ K ⁡ W
dihjust.u ⊢ U = DVecH ⁡ K ⁡ W
dihjust.s ⊢ ⊕ ˙ = LSSum ⁡ U
Assertion dihjust ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W = J ⁡ R ⊕ ˙ I ⁡ X ∧ ˙ W

Proof

Step Hyp Ref Expression
1 dihjust.b ⊢ B = Base K
2 dihjust.l ⊢ ≤ ˙ = ≤ K
3 dihjust.j ⊢ ∨ ˙ = join ⁡ K
4 dihjust.m ⊢ ∧ ˙ = meet ⁡ K
5 dihjust.a ⊢ A = Atoms ⁡ K
6 dihjust.h ⊢ H = LHyp ⁡ K
7 dihjust.i ⊢ I = DIsoB ⁡ K ⁡ W
8 dihjust.J ⊢ J = DIsoC ⁡ K ⁡ W
9 dihjust.u ⊢ U = DVecH ⁡ K ⁡ W
10 dihjust.s ⊢ ⊕ ˙ = LSSum ⁡ U
11 1 2 3 4 5 6 7 8 9 10 dihjustlem ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ R ⊕ ˙ I ⁡ X ∧ ˙ W
12 simp1 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → K ∈ HL ∧ W ∈ H
13 simp22 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → R ∈ A ∧ ¬ R ≤ ˙ W
14 simp21 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → Q ∈ A ∧ ¬ Q ≤ ˙ W
15 simp23 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → X ∈ B
16 simp3 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W
17 16 eqcomd ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → R ∨ ˙ X ∧ ˙ W = Q ∨ ˙ X ∧ ˙ W
18 1 2 3 4 5 6 7 8 9 10 dihjustlem ⊢ K ∈ HL ∧ W ∈ H ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ X ∈ B ∧ R ∨ ˙ X ∧ ˙ W = Q ∨ ˙ X ∧ ˙ W → J ⁡ R ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W
19 12 13 14 15 17 18 syl131anc ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → J ⁡ R ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W
20 11 19 eqssd ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ X ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = R ∨ ˙ X ∧ ˙ W → J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W = J ⁡ R ⊕ ˙ I ⁡ X ∧ ˙ W