Metamath Proof Explorer


Theorem 4atex2-0cOLDN

Description: Same as 4atex2 except that S and T are zero. TODO: do we need this one or 4atex2-0aOLDN or 4atex2-0bOLDN ? (Contributed by NM, 27-May-2013) (New usage is discouraged.)

Ref Expression
Hypotheses 4that.l ⊢ ≤ ˙ = ≤ K
4that.j ⊢ ∨ ˙ = join ⁡ K
4that.a ⊢ A = Atoms ⁡ K
4that.h ⊢ H = LHyp ⁡ K
Assertion 4atex2-0cOLDN ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ S ∨ ˙ z = T ∨ ˙ z

Proof

Step Hyp Ref Expression
1 4that.l ⊢ ≤ ˙ = ≤ K
2 4that.j ⊢ ∨ ˙ = join ⁡ K
3 4that.a ⊢ A = Atoms ⁡ K
4 4that.h ⊢ H = LHyp ⁡ K
5 simp21l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → P ∈ A
6 simp21r ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ¬ P ≤ ˙ W
7 simp23 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → S = 0. ⁡ K
8 7 oveq1d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → S ∨ ˙ P = 0. ⁡ K ∨ ˙ P
9 simp32 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → T = 0. ⁡ K
10 9 oveq1d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → T ∨ ˙ P = 0. ⁡ K ∨ ˙ P
11 8 10 eqtr4d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → S ∨ ˙ P = T ∨ ˙ P
12 breq1 ⊢ z = P → z ≤ ˙ W ↔ P ≤ ˙ W
13 12 notbid ⊢ z = P → ¬ z ≤ ˙ W ↔ ¬ P ≤ ˙ W
14 oveq2 ⊢ z = P → S ∨ ˙ z = S ∨ ˙ P
15 oveq2 ⊢ z = P → T ∨ ˙ z = T ∨ ˙ P
16 14 15 eqeq12d ⊢ z = P → S ∨ ˙ z = T ∨ ˙ z ↔ S ∨ ˙ P = T ∨ ˙ P
17 13 16 anbi12d ⊢ z = P → ¬ z ≤ ˙ W ∧ S ∨ ˙ z = T ∨ ˙ z ↔ ¬ P ≤ ˙ W ∧ S ∨ ˙ P = T ∨ ˙ P
18 17 rspcev ⊢ P ∈ A ∧ ¬ P ≤ ˙ W ∧ S ∨ ˙ P = T ∨ ˙ P → ∃ z ∈ A ¬ z ≤ ˙ W ∧ S ∨ ˙ z = T ∨ ˙ z
19 5 6 11 18 syl12anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S = 0. ⁡ K ∧ P ≠ Q ∧ T = 0. ⁡ K ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ S ∨ ˙ z = T ∨ ˙ z