Description: Part of proof of Lemma K of Crawley p. 118. Line 16 on p. 119 for i = 2, where sigma_2 (p) is V , f_2 is C , and k_2 is Q . (Contributed by NM, 2-Jul-2013)
Ref | Expression | ||
---|---|---|---|
Hypotheses | cdlemk2.b | ⊢ 𝐵 = ( Base ‘ 𝐾 ) | |
cdlemk2.l | ⊢ ≤ = ( le ‘ 𝐾 ) | ||
cdlemk2.j | ⊢ ∨ = ( join ‘ 𝐾 ) | ||
cdlemk2.m | ⊢ ∧ = ( meet ‘ 𝐾 ) | ||
cdlemk2.a | ⊢ 𝐴 = ( Atoms ‘ 𝐾 ) | ||
cdlemk2.h | ⊢ 𝐻 = ( LHyp ‘ 𝐾 ) | ||
cdlemk2.t | ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 ) | ||
cdlemk2.r | ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 ) | ||
cdlemk2.s | ⊢ 𝑆 = ( 𝑓 ∈ 𝑇 ↦ ( ℩ 𝑖 ∈ 𝑇 ( 𝑖 ‘ 𝑃 ) = ( ( 𝑃 ∨ ( 𝑅 ‘ 𝑓 ) ) ∧ ( ( 𝑁 ‘ 𝑃 ) ∨ ( 𝑅 ‘ ( 𝑓 ∘ ◡ 𝐹 ) ) ) ) ) ) | ||
cdlemk2.q | ⊢ 𝑄 = ( 𝑆 ‘ 𝐶 ) | ||
cdlemk2.v | ⊢ 𝑉 = ( 𝑑 ∈ 𝑇 ↦ ( ℩ 𝑘 ∈ 𝑇 ( 𝑘 ‘ 𝑃 ) = ( ( 𝑃 ∨ ( 𝑅 ‘ 𝑑 ) ) ∧ ( ( 𝑄 ‘ 𝑃 ) ∨ ( 𝑅 ‘ ( 𝑑 ∘ ◡ 𝐶 ) ) ) ) ) ) | ||
Assertion | cdlemkuv2-2 | ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( ( 𝑅 ‘ 𝐶 ) ≠ ( 𝑅 ‘ 𝐹 ) ∧ ( 𝑅 ‘ 𝐶 ) ≠ ( 𝑅 ‘ 𝐺 ) ) ∧ ( 𝐹 ≠ ( I ↾ 𝐵 ) ∧ 𝐺 ≠ ( I ↾ 𝐵 ) ∧ 𝐶 ≠ ( I ↾ 𝐵 ) ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( ( 𝑉 ‘ 𝐺 ) ‘ 𝑃 ) = ( ( 𝑃 ∨ ( 𝑅 ‘ 𝐺 ) ) ∧ ( ( 𝑄 ‘ 𝑃 ) ∨ ( 𝑅 ‘ ( 𝐺 ∘ ◡ 𝐶 ) ) ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | cdlemk2.b | ⊢ 𝐵 = ( Base ‘ 𝐾 ) | |
2 | cdlemk2.l | ⊢ ≤ = ( le ‘ 𝐾 ) | |
3 | cdlemk2.j | ⊢ ∨ = ( join ‘ 𝐾 ) | |
4 | cdlemk2.m | ⊢ ∧ = ( meet ‘ 𝐾 ) | |
5 | cdlemk2.a | ⊢ 𝐴 = ( Atoms ‘ 𝐾 ) | |
6 | cdlemk2.h | ⊢ 𝐻 = ( LHyp ‘ 𝐾 ) | |
7 | cdlemk2.t | ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 ) | |
8 | cdlemk2.r | ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 ) | |
9 | cdlemk2.s | ⊢ 𝑆 = ( 𝑓 ∈ 𝑇 ↦ ( ℩ 𝑖 ∈ 𝑇 ( 𝑖 ‘ 𝑃 ) = ( ( 𝑃 ∨ ( 𝑅 ‘ 𝑓 ) ) ∧ ( ( 𝑁 ‘ 𝑃 ) ∨ ( 𝑅 ‘ ( 𝑓 ∘ ◡ 𝐹 ) ) ) ) ) ) | |
10 | cdlemk2.q | ⊢ 𝑄 = ( 𝑆 ‘ 𝐶 ) | |
11 | cdlemk2.v | ⊢ 𝑉 = ( 𝑑 ∈ 𝑇 ↦ ( ℩ 𝑘 ∈ 𝑇 ( 𝑘 ‘ 𝑃 ) = ( ( 𝑃 ∨ ( 𝑅 ‘ 𝑑 ) ) ∧ ( ( 𝑄 ‘ 𝑃 ) ∨ ( 𝑅 ‘ ( 𝑑 ∘ ◡ 𝐶 ) ) ) ) ) ) | |
12 | 1 2 3 4 5 6 7 8 9 10 11 | cdlemkuv2 | ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ 𝐺 ∈ 𝑇 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( ( 𝑅 ‘ 𝐶 ) ≠ ( 𝑅 ‘ 𝐹 ) ∧ ( 𝑅 ‘ 𝐶 ) ≠ ( 𝑅 ‘ 𝐺 ) ) ∧ ( 𝐹 ≠ ( I ↾ 𝐵 ) ∧ 𝐺 ≠ ( I ↾ 𝐵 ) ∧ 𝐶 ≠ ( I ↾ 𝐵 ) ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( ( 𝑉 ‘ 𝐺 ) ‘ 𝑃 ) = ( ( 𝑃 ∨ ( 𝑅 ‘ 𝐺 ) ) ∧ ( ( 𝑄 ‘ 𝑃 ) ∨ ( 𝑅 ‘ ( 𝐺 ∘ ◡ 𝐶 ) ) ) ) ) |