Metamath Proof Explorer


Theorem isoml

Description: The predicate "is an orthomodular lattice." (Contributed by NM, 18-Sep-2011)

Ref Expression
Hypotheses isoml.b ⊢ B = Base K
isoml.l ⊢ ≤ ˙ = ≤ K
isoml.j ⊢ ∨ ˙ = join ⁡ K
isoml.m ⊢ ∧ ˙ = meet ⁡ K
isoml.o ⊢ ⊥ ˙ = oc ⁡ K
Assertion isoml ⊢ K ∈ OML ↔ K ∈ OL ∧ ∀ x ∈ B ∀ y ∈ B x ≤ ˙ y → y = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x

Proof

Step Hyp Ref Expression
1 isoml.b ⊢ B = Base K
2 isoml.l ⊢ ≤ ˙ = ≤ K
3 isoml.j ⊢ ∨ ˙ = join ⁡ K
4 isoml.m ⊢ ∧ ˙ = meet ⁡ K
5 isoml.o ⊢ ⊥ ˙ = oc ⁡ K
6 fveq2 ⊢ k = K → Base k = Base K
7 6 1 eqtr4di ⊢ k = K → Base k = B
8 fveq2 ⊢ k = K → ≤ k = ≤ K
9 8 2 eqtr4di ⊢ k = K → ≤ k = ≤ ˙
10 9 breqd ⊢ k = K → x ≤ k y ↔ x ≤ ˙ y
11 fveq2 ⊢ k = K → join ⁡ k = join ⁡ K
12 11 3 eqtr4di ⊢ k = K → join ⁡ k = ∨ ˙
13 eqidd ⊢ k = K → x = x
14 fveq2 ⊢ k = K → meet ⁡ k = meet ⁡ K
15 14 4 eqtr4di ⊢ k = K → meet ⁡ k = ∧ ˙
16 eqidd ⊢ k = K → y = y
17 fveq2 ⊢ k = K → oc ⁡ k = oc ⁡ K
18 17 5 eqtr4di ⊢ k = K → oc ⁡ k = ⊥ ˙
19 18 fveq1d ⊢ k = K → oc ⁡ k ⁡ x = ⊥ ˙ ⁡ x
20 15 16 19 oveq123d ⊢ k = K → y meet ⁡ k oc ⁡ k ⁡ x = y ∧ ˙ ⊥ ˙ ⁡ x
21 12 13 20 oveq123d ⊢ k = K → x join ⁡ k y meet ⁡ k oc ⁡ k ⁡ x = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x
22 21 eqeq2d ⊢ k = K → y = x join ⁡ k y meet ⁡ k oc ⁡ k ⁡ x ↔ y = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x
23 10 22 imbi12d ⊢ k = K → x ≤ k y → y = x join ⁡ k y meet ⁡ k oc ⁡ k ⁡ x ↔ x ≤ ˙ y → y = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x
24 7 23 raleqbidv ⊢ k = K → ∀ y ∈ Base k x ≤ k y → y = x join ⁡ k y meet ⁡ k oc ⁡ k ⁡ x ↔ ∀ y ∈ B x ≤ ˙ y → y = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x
25 7 24 raleqbidv ⊢ k = K → ∀ x ∈ Base k ∀ y ∈ Base k x ≤ k y → y = x join ⁡ k y meet ⁡ k oc ⁡ k ⁡ x ↔ ∀ x ∈ B ∀ y ∈ B x ≤ ˙ y → y = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x
26 df-oml ⊢ OML = k ∈ OL | ∀ x ∈ Base k ∀ y ∈ Base k x ≤ k y → y = x join ⁡ k y meet ⁡ k oc ⁡ k ⁡ x
27 25 26 elrab2 ⊢ K ∈ OML ↔ K ∈ OL ∧ ∀ x ∈ B ∀ y ∈ B x ≤ ˙ y → y = x ∨ ˙ y ∧ ˙ ⊥ ˙ ⁡ x