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Metamath Proof Explorer

Theorem doch11

Description: Orthocomplement is one-to-one. (Contributed by NM, 12-Aug-2014)

Ref Expression
Hypotheses doch11.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
doch11.i ⊢ 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
doch11.o ⊢ ⊥ = ( ( ocH ‘ 𝐾 ) ‘ 𝑊 )
doch11.k ⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
doch11.x ⊢ ( 𝜑 → 𝑋 ∈ ran 𝐼 )
doch11.y ⊢ ( 𝜑 → 𝑌 ∈ ran 𝐼 )
Assertion doch11 ( 𝜑 → ( ( ⊥ ‘ 𝑋 ) = ( ⊥ ‘ 𝑌 ) ↔ 𝑋 = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 doch11.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
2 doch11.i ⊢ 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
3 doch11.o ⊢ ⊥ = ( ( ocH ‘ 𝐾 ) ‘ 𝑊 )
4 doch11.k ⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
5 doch11.x ⊢ ( 𝜑 → 𝑋 ∈ ran 𝐼 )
6 doch11.y ⊢ ( 𝜑 → 𝑌 ∈ ran 𝐼 )
7 1 2 3 4 6 5 dochord ⊢ ( 𝜑 → ( 𝑌 ⊆ 𝑋 ↔ ( ⊥ ‘ 𝑋 ) ⊆ ( ⊥ ‘ 𝑌 ) ) )
8 1 2 3 4 5 6 dochord ⊢ ( 𝜑 → ( 𝑋 ⊆ 𝑌 ↔ ( ⊥ ‘ 𝑌 ) ⊆ ( ⊥ ‘ 𝑋 ) ) )
9 7 8 anbi12d ⊢ ( 𝜑 → ( ( 𝑌 ⊆ 𝑋 ∧ 𝑋 ⊆ 𝑌 ) ↔ ( ( ⊥ ‘ 𝑋 ) ⊆ ( ⊥ ‘ 𝑌 ) ∧ ( ⊥ ‘ 𝑌 ) ⊆ ( ⊥ ‘ 𝑋 ) ) ) )
10 eqcom ⊢ ( 𝑋 = 𝑌 ↔ 𝑌 = 𝑋 )
11 eqss ⊢ ( 𝑌 = 𝑋 ↔ ( 𝑌 ⊆ 𝑋 ∧ 𝑋 ⊆ 𝑌 ) )
12 10 11 bitri ⊢ ( 𝑋 = 𝑌 ↔ ( 𝑌 ⊆ 𝑋 ∧ 𝑋 ⊆ 𝑌 ) )
13 eqss ⊢ ( ( ⊥ ‘ 𝑋 ) = ( ⊥ ‘ 𝑌 ) ↔ ( ( ⊥ ‘ 𝑋 ) ⊆ ( ⊥ ‘ 𝑌 ) ∧ ( ⊥ ‘ 𝑌 ) ⊆ ( ⊥ ‘ 𝑋 ) ) )
14 9 12 13 3bitr4g ⊢ ( 𝜑 → ( 𝑋 = 𝑌 ↔ ( ⊥ ‘ 𝑋 ) = ( ⊥ ‘ 𝑌 ) ) )
15 14 bicomd ⊢ ( 𝜑 → ( ( ⊥ ‘ 𝑋 ) = ( ⊥ ‘ 𝑌 ) ↔ 𝑋 = 𝑌 ) )