Metamath Proof Explorer


Theorem 4atexlem7

Description: Whenever there are at least 4 atoms under P .\/ Q (specifically, P , Q , r , and ( P .\/ Q ) ./\ W ), there are also at least 4 atoms under P .\/ S . This proves the statement in Lemma E of Crawley p. 114, last line, "...p \/ q/0 and hence p \/ s/0 contains at least four atoms..." Note that by cvlsupr2 , our ( P .\/ r ) = ( Q .\/ r ) is a shorter way to express r =/= P /\ r =/= Q /\ r .<_ ( P .\/ Q ) . With a longer proof, the condition -. S .<_ ( P .\/ Q ) could be eliminated (see 4atex ), although for some purposes this more restricted lemma may be adequate. (Contributed by NM, 25-Nov-2012)

Ref Expression
Hypotheses 4that.l ⊢ ≤ ˙ = ≤ K
4that.j ⊢ ∨ ˙ = join ⁡ K
4that.a ⊢ A = Atoms ⁡ K
4that.h ⊢ H = LHyp ⁡ K
Assertion 4atexlem7 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ 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 simp11l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → K ∈ HL ∧ W ∈ H
6 simp1r1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → P ∈ A ∧ ¬ P ≤ ˙ W
7 6 3ad2ant1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → P ∈ A ∧ ¬ P ≤ ˙ W
8 simp1r2 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ∈ A ∧ ¬ Q ≤ ˙ W
9 8 3ad2ant1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → Q ∈ A ∧ ¬ Q ≤ ˙ W
10 simp2 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → r ∈ A
11 simp3l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ¬ r ≤ ˙ W
12 10 11 jca ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → r ∈ A ∧ ¬ r ≤ ˙ W
13 simp1r3 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → S ∈ A
14 13 3ad2ant1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → S ∈ A
15 simp3r ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → P ∨ ˙ r = Q ∨ ˙ r
16 simp12 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → P ≠ Q
17 simp13 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ¬ S ≤ ˙ P ∨ ˙ Q
18 eqid ⊢ meet ⁡ K = meet ⁡ K
19 1 2 18 3 4 4atexlemex6 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ S ∈ A ∧ P ∨ ˙ r = Q ∨ ˙ r ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ z
20 5 7 9 12 14 15 16 17 19 syl323anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ r ∈ A ∧ ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ z
21 20 rexlimdv3a ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ z
22 21 3exp ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A → P ≠ Q → ¬ S ≤ ˙ P ∨ ˙ Q → ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ z
23 22 3impd ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A → P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ z
24 23 3impia ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ S ∈ A ∧ P ≠ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q ∧ ∃ r ∈ A ¬ r ≤ ˙ W ∧ P ∨ ˙ r = Q ∨ ˙ r → ∃ z ∈ A ¬ z ≤ ˙ W ∧ P ∨ ˙ z = S ∨ ˙ z