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