Metamath Proof Explorer


Theorem cdleml5N

Description: Part of proof of Lemma L of Crawley p. 120. TODO: fix comment. (Contributed by NM, 1-Aug-2013) (New usage is discouraged.)

Ref Expression
Hypotheses cdleml1.b ⊢ B = Base K
cdleml1.h ⊢ H = LHyp ⁡ K
cdleml1.t ⊢ T = LTrn ⁡ K ⁡ W
cdleml1.r ⊢ R = trL ⁡ K ⁡ W
cdleml1.e ⊢ E = TEndo ⁡ K ⁡ W
cdleml3.o ⊢ 0 ˙ = g ∈ T ⟼ I ↾ B
Assertion cdleml5N ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ → ∃ s ∈ E s ∘ U = V

Proof

Step Hyp Ref Expression
1 cdleml1.b ⊢ B = Base K
2 cdleml1.h ⊢ H = LHyp ⁡ K
3 cdleml1.t ⊢ T = LTrn ⁡ K ⁡ W
4 cdleml1.r ⊢ R = trL ⁡ K ⁡ W
5 cdleml1.e ⊢ E = TEndo ⁡ K ⁡ W
6 cdleml3.o ⊢ 0 ˙ = g ∈ T ⟼ I ↾ B
7 simpl1 ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → K ∈ HL ∧ W ∈ H
8 1 2 3 5 6 tendo0cl ⊢ K ∈ HL ∧ W ∈ H → 0 ˙ ∈ E
9 7 8 syl ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → 0 ˙ ∈ E
10 simpl2l ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → U ∈ E
11 1 2 3 5 6 tendo0mul ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E → 0 ˙ ∘ U = 0 ˙
12 7 10 11 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → 0 ˙ ∘ U = 0 ˙
13 simpr ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → V = 0 ˙
14 12 13 eqtr4d ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → 0 ˙ ∘ U = V
15 coeq1 ⊢ s = 0 ˙ → s ∘ U = 0 ˙ ∘ U
16 15 eqeq1d ⊢ s = 0 ˙ → s ∘ U = V ↔ 0 ˙ ∘ U = V
17 16 rspcev ⊢ 0 ˙ ∈ E ∧ 0 ˙ ∘ U = V → ∃ s ∈ E s ∘ U = V
18 9 14 17 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V = 0 ˙ → ∃ s ∈ E s ∘ U = V
19 simpl1 ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V ≠ 0 ˙ → K ∈ HL ∧ W ∈ H
20 simpl2 ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V ≠ 0 ˙ → U ∈ E ∧ V ∈ E
21 simpl3 ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V ≠ 0 ˙ → U ≠ 0 ˙
22 simpr ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V ≠ 0 ˙ → V ≠ 0 ˙
23 1 2 3 4 5 6 cdleml4N ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V ≠ 0 ˙ → ∃ s ∈ E s ∘ U = V
24 19 20 21 22 23 syl112anc ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ ∧ V ≠ 0 ˙ → ∃ s ∈ E s ∘ U = V
25 18 24 pm2.61dane ⊢ K ∈ HL ∧ W ∈ H ∧ U ∈ E ∧ V ∈ E ∧ U ≠ 0 ˙ → ∃ s ∈ E s ∘ U = V