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Metamath Proof Explorer

Theorem pjoml2

Description: Variation of orthomodular law. Definition in Kalmbach p. 22. (Contributed by NM, 13-Jun-2006) (New usage is discouraged.)

Ref Expression
Assertion pjoml2 ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ A ⊆ B → A ∨ ℋ ⊥ ⁡ A ∩ B = B

Proof

Step Hyp Ref Expression
1 sseq1 ⊢ A = if A ∈ C ℋ A 0 ℋ → A ⊆ B ↔ if A ∈ C ℋ A 0 ℋ ⊆ B
2 id ⊢ A = if A ∈ C ℋ A 0 ℋ → A = if A ∈ C ℋ A 0 ℋ
3 fveq2 ⊢ A = if A ∈ C ℋ A 0 ℋ → ⊥ ⁡ A = ⊥ ⁡ if A ∈ C ℋ A 0 ℋ
4 3 ineq1d ⊢ A = if A ∈ C ℋ A 0 ℋ → ⊥ ⁡ A ∩ B = ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B
5 2 4 oveq12d ⊢ A = if A ∈ C ℋ A 0 ℋ → A ∨ ℋ ⊥ ⁡ A ∩ B = if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B
6 5 eqeq1d ⊢ A = if A ∈ C ℋ A 0 ℋ → A ∨ ℋ ⊥ ⁡ A ∩ B = B ↔ if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B = B
7 1 6 imbi12d ⊢ A = if A ∈ C ℋ A 0 ℋ → A ⊆ B → A ∨ ℋ ⊥ ⁡ A ∩ B = B ↔ if A ∈ C ℋ A 0 ℋ ⊆ B → if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B = B
8 sseq2 ⊢ B = if B ∈ C ℋ B 0 ℋ → if A ∈ C ℋ A 0 ℋ ⊆ B ↔ if A ∈ C ℋ A 0 ℋ ⊆ if B ∈ C ℋ B 0 ℋ
9 ineq2 ⊢ B = if B ∈ C ℋ B 0 ℋ → ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B = ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ if B ∈ C ℋ B 0 ℋ
10 9 oveq2d ⊢ B = if B ∈ C ℋ B 0 ℋ → if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B = if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ if B ∈ C ℋ B 0 ℋ
11 id ⊢ B = if B ∈ C ℋ B 0 ℋ → B = if B ∈ C ℋ B 0 ℋ
12 10 11 eqeq12d ⊢ B = if B ∈ C ℋ B 0 ℋ → if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B = B ↔ if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ if B ∈ C ℋ B 0 ℋ = if B ∈ C ℋ B 0 ℋ
13 8 12 imbi12d ⊢ B = if B ∈ C ℋ B 0 ℋ → if A ∈ C ℋ A 0 ℋ ⊆ B → if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ B = B ↔ if A ∈ C ℋ A 0 ℋ ⊆ if B ∈ C ℋ B 0 ℋ → if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ if B ∈ C ℋ B 0 ℋ = if B ∈ C ℋ B 0 ℋ
14 h0elch ⊢ 0 ℋ ∈ C ℋ
15 14 elimel ⊢ if A ∈ C ℋ A 0 ℋ ∈ C ℋ
16 14 elimel ⊢ if B ∈ C ℋ B 0 ℋ ∈ C ℋ
17 15 16 pjoml2i ⊢ if A ∈ C ℋ A 0 ℋ ⊆ if B ∈ C ℋ B 0 ℋ → if A ∈ C ℋ A 0 ℋ ∨ ℋ ⊥ ⁡ if A ∈ C ℋ A 0 ℋ ∩ if B ∈ C ℋ B 0 ℋ = if B ∈ C ℋ B 0 ℋ
18 7 13 17 dedth2h ⊢ A ∈ C ℋ ∧ B ∈ C ℋ → A ⊆ B → A ∨ ℋ ⊥ ⁡ A ∩ B = B
19 18 3impia ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ A ⊆ B → A ∨ ℋ ⊥ ⁡ A ∩ B = B