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Metamath Proof Explorer

Theorem cvlsupr3

Description: Two equivalent ways of expressing that R is a superposition of P and Q , which can replace the superposition part of ishlat1 , ( x =/= y -> E. z e. A ( z =/= x /\ z =/= y /\ z .<_ ( x .\/ y ) ) ) , with the simpler E. z e. A ( x .\/ z ) = ( y .\/ z ) as shown in ishlat3N . (Contributed by NM, 5-Nov-2012)

Ref Expression
Hypotheses cvlsupr2.a ⊢ A = Atoms ⁡ K
cvlsupr2.l ⊢ ≤ ˙ = ≤ K
cvlsupr2.j ⊢ ∨ ˙ = join ⁡ K
Assertion cvlsupr3 ⊢ K ∈ CvLat ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A → P ∨ ˙ R = Q ∨ ˙ R ↔ P ≠ Q → R ≠ P ∧ R ≠ Q ∧ R ≤ ˙ P ∨ ˙ Q

Proof

Step Hyp Ref Expression
1 cvlsupr2.a ⊢ A = Atoms ⁡ K
2 cvlsupr2.l ⊢ ≤ ˙ = ≤ K
3 cvlsupr2.j ⊢ ∨ ˙ = join ⁡ K
4 df-ne ⊢ P ≠ Q ↔ ¬ P = Q
5 4 imbi1i ⊢ P ≠ Q → P ∨ ˙ R = Q ∨ ˙ R ↔ ¬ P = Q → P ∨ ˙ R = Q ∨ ˙ R
6 oveq1 ⊢ P = Q → P ∨ ˙ R = Q ∨ ˙ R
7 6 biantrur ⊢ ¬ P = Q → P ∨ ˙ R = Q ∨ ˙ R ↔ P = Q → P ∨ ˙ R = Q ∨ ˙ R ∧ ¬ P = Q → P ∨ ˙ R = Q ∨ ˙ R
8 pm4.83 ⊢ P = Q → P ∨ ˙ R = Q ∨ ˙ R ∧ ¬ P = Q → P ∨ ˙ R = Q ∨ ˙ R ↔ P ∨ ˙ R = Q ∨ ˙ R
9 5 7 8 3bitrri ⊢ P ∨ ˙ R = Q ∨ ˙ R ↔ P ≠ Q → P ∨ ˙ R = Q ∨ ˙ R
10 1 2 3 cvlsupr2 ⊢ K ∈ CvLat ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ Q → P ∨ ˙ R = Q ∨ ˙ R ↔ R ≠ P ∧ R ≠ Q ∧ R ≤ ˙ P ∨ ˙ Q
11 10 3expia ⊢ K ∈ CvLat ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A → P ≠ Q → P ∨ ˙ R = Q ∨ ˙ R ↔ R ≠ P ∧ R ≠ Q ∧ R ≤ ˙ P ∨ ˙ Q
12 11 pm5.74d ⊢ K ∈ CvLat ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A → P ≠ Q → P ∨ ˙ R = Q ∨ ˙ R ↔ P ≠ Q → R ≠ P ∧ R ≠ Q ∧ R ≤ ˙ P ∨ ˙ Q
13 9 12 bitrid ⊢ K ∈ CvLat ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A → P ∨ ˙ R = Q ∨ ˙ R ↔ P ≠ Q → R ≠ P ∧ R ≠ Q ∧ R ≤ ˙ P ∨ ˙ Q