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Metamath Proof Explorer

Theorem pjoml

Description: Subspace form of orthomodular law in the Hilbert lattice. Compare the orthomodular law in Theorem 2(ii) of Kalmbach p. 22. Derived using projections; compare omlsi . (Contributed by NM, 14-Jun-2006) (New usage is discouraged.)

Ref Expression
Assertion pjoml ( ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Sℋ ) ∧ ( 𝐴 ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = 0ℋ ) ) → 𝐴 = 𝐵 )

Proof

Step Hyp Ref Expression
1 sseq1 ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( 𝐴 ⊆ 𝐵 ↔ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ) )
2 fveq2 ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ⊥ ‘ 𝐴 ) = ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) )
3 2 ineq2d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) )
4 3 eqeq1d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = 0ℋ ↔ ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) )
5 1 4 anbi12d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ( 𝐴 ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = 0ℋ ) ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) ) )
6 eqeq1 ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( 𝐴 = 𝐵 ↔ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = 𝐵 ) )
7 5 6 imbi12d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ( ( 𝐴 ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = 0ℋ ) → 𝐴 = 𝐵 ) ↔ ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) → if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = 𝐵 ) ) )
8 sseq2 ⊢ ( 𝐵 = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) → ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ↔ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ) )
9 ineq1 ⊢ ( 𝐵 = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) → ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = ( if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) )
10 9 eqeq1d ⊢ ( 𝐵 = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) → ( ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ↔ ( if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) )
11 8 10 anbi12d ⊢ ( 𝐵 = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) → ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∧ ( if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) ) )
12 eqeq2 ⊢ ( 𝐵 = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) → ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = 𝐵 ↔ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ) )
13 11 12 imbi12d ⊢ ( 𝐵 = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) → ( ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) → if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = 𝐵 ) ↔ ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∧ ( if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) → if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ) ) )
14 h0elch ⊢ 0ℋ ∈ Cℋ
15 14 elimel ⊢ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∈ Cℋ
16 h0elsh ⊢ 0ℋ ∈ Sℋ
17 16 elimel ⊢ if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∈ Sℋ
18 15 17 pjomli ⊢ ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∧ ( if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) ∩ ( ⊥ ‘ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) ) = 0ℋ ) → if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) = if ( 𝐵 ∈ Sℋ , 𝐵 , 0ℋ ) )
19 7 13 18 dedth2h ⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐴 ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = 0ℋ ) → 𝐴 = 𝐵 ) )
20 19 imp ⊢ ( ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Sℋ ) ∧ ( 𝐴 ⊆ 𝐵 ∧ ( 𝐵 ∩ ( ⊥ ‘ 𝐴 ) ) = 0ℋ ) ) → 𝐴 = 𝐵 )