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Metamath Proof Explorer

Theorem dihcnv11

Description: The converse of isomorphism H is one-to-one. (Contributed by NM, 17-Aug-2014)

Ref Expression
Hypotheses dihcnv11.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
dihcnv11.i ⊢ 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
dihcnv11.k ⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
dihcnv11.x ⊢ ( 𝜑 → 𝑋 ∈ ran 𝐼 )
dihcnv11.y ⊢ ( 𝜑 → 𝑌 ∈ ran 𝐼 )
Assertion dihcnv11 ( 𝜑 → ( ( ◡ 𝐼 ‘ 𝑋 ) = ( ◡ 𝐼 ‘ 𝑌 ) ↔ 𝑋 = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 dihcnv11.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
2 dihcnv11.i ⊢ 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
3 dihcnv11.k ⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
4 dihcnv11.x ⊢ ( 𝜑 → 𝑋 ∈ ran 𝐼 )
5 dihcnv11.y ⊢ ( 𝜑 → 𝑌 ∈ ran 𝐼 )
6 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
7 6 1 2 dihcnvcl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ ran 𝐼 ) → ( ◡ 𝐼 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) )
8 3 4 7 syl2anc ⊢ ( 𝜑 → ( ◡ 𝐼 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) )
9 6 1 2 dihcnvcl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑌 ∈ ran 𝐼 ) → ( ◡ 𝐼 ‘ 𝑌 ) ∈ ( Base ‘ 𝐾 ) )
10 3 5 9 syl2anc ⊢ ( 𝜑 → ( ◡ 𝐼 ‘ 𝑌 ) ∈ ( Base ‘ 𝐾 ) )
11 6 1 2 dih11 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ◡ 𝐼 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) ∧ ( ◡ 𝐼 ‘ 𝑌 ) ∈ ( Base ‘ 𝐾 ) ) → ( ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑋 ) ) = ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑌 ) ) ↔ ( ◡ 𝐼 ‘ 𝑋 ) = ( ◡ 𝐼 ‘ 𝑌 ) ) )
12 3 8 10 11 syl3anc ⊢ ( 𝜑 → ( ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑋 ) ) = ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑌 ) ) ↔ ( ◡ 𝐼 ‘ 𝑋 ) = ( ◡ 𝐼 ‘ 𝑌 ) ) )
13 1 2 dihcnvid2 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑋 ∈ ran 𝐼 ) → ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑋 ) ) = 𝑋 )
14 3 4 13 syl2anc ⊢ ( 𝜑 → ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑋 ) ) = 𝑋 )
15 1 2 dihcnvid2 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑌 ∈ ran 𝐼 ) → ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑌 ) ) = 𝑌 )
16 3 5 15 syl2anc ⊢ ( 𝜑 → ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑌 ) ) = 𝑌 )
17 14 16 eqeq12d ⊢ ( 𝜑 → ( ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑋 ) ) = ( 𝐼 ‘ ( ◡ 𝐼 ‘ 𝑌 ) ) ↔ 𝑋 = 𝑌 ) )
18 12 17 bitr3d ⊢ ( 𝜑 → ( ( ◡ 𝐼 ‘ 𝑋 ) = ( ◡ 𝐼 ‘ 𝑌 ) ↔ 𝑋 = 𝑌 ) )