Metamath Proof Explorer


Theorem cdleme19f

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