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Metamath Proof Explorer

Theorem trlconid

Description: The composition of two different translations is not the identity translation. (Contributed by NM, 22-Jul-2013)

Ref Expression
Hypotheses trlconid.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
trlconid.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
trlconid.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
trlconid.r ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 )
Assertion trlconid ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → ( 𝐹 ∘ 𝐺 ) ≠ ( I ↾ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 trlconid.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 trlconid.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
3 trlconid.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
4 trlconid.r ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 )
5 eqid ⊢ ( Atoms ‘ 𝐾 ) = ( Atoms ‘ 𝐾 )
6 5 2 3 4 trlcoat ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ∈ ( Atoms ‘ 𝐾 ) )
7 simp1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
8 simp2l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → 𝐹 ∈ 𝑇 )
9 simp2r ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → 𝐺 ∈ 𝑇 )
10 2 3 ltrnco ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( 𝐹 ∘ 𝐺 ) ∈ 𝑇 )
11 7 8 9 10 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → ( 𝐹 ∘ 𝐺 ) ∈ 𝑇 )
12 1 5 2 3 4 trlnidatb ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∘ 𝐺 ) ∈ 𝑇 ) → ( ( 𝐹 ∘ 𝐺 ) ≠ ( I ↾ 𝐵 ) ↔ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ∈ ( Atoms ‘ 𝐾 ) ) )
13 7 11 12 syl2anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → ( ( 𝐹 ∘ 𝐺 ) ≠ ( I ↾ 𝐵 ) ↔ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ∈ ( Atoms ‘ 𝐾 ) ) )
14 6 13 mpbird ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝑅 ‘ 𝐹 ) ≠ ( 𝑅 ‘ 𝐺 ) ) → ( 𝐹 ∘ 𝐺 ) ≠ ( I ↾ 𝐵 ) )