Metamath Proof Explorer

Theorem cdlemk1

Description: Part of proof of Lemma K of Crawley p. 118. (Contributed by NM, 22-Jun-2013)

Ref Expression
Hypotheses cdlemk.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
cdlemk.l ⊢ ≤ = ( le ‘ 𝐾 )
cdlemk.j ⊢ ∨ = ( join ‘ 𝐾 )
cdlemk.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
cdlemk.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
cdlemk.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
cdlemk.r ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 )
Assertion cdlemk1 ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑃 ∨ ( 𝑁 ‘ 𝑃 ) ) = ( ( 𝐹 ‘ 𝑃 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )

Proof

Step Hyp Ref Expression
1 cdlemk.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 cdlemk.l ⊢ ≤ = ( le ‘ 𝐾 )
3 cdlemk.j ⊢ ∨ = ( join ‘ 𝐾 )
4 cdlemk.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
5 cdlemk.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
6 cdlemk.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
7 cdlemk.r ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 )
8 simp3l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) )
9 8 oveq2d ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑃 ∨ ( 𝑅 ‘ 𝐹 ) ) = ( 𝑃 ∨ ( 𝑅 ‘ 𝑁 ) ) )
10 simp1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
11 simp2l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → 𝐹 ∈ 𝑇 )
12 simp3r ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) )
13 2 3 4 5 6 7 trljat3 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) → ( 𝑃 ∨ ( 𝑅 ‘ 𝐹 ) ) = ( ( 𝐹 ‘ 𝑃 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )
14 10 11 12 13 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑃 ∨ ( 𝑅 ‘ 𝐹 ) ) = ( ( 𝐹 ‘ 𝑃 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )
15 simp2r ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → 𝑁 ∈ 𝑇 )
16 2 3 4 5 6 7 trljat1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑁 ∈ 𝑇 ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) → ( 𝑃 ∨ ( 𝑅 ‘ 𝑁 ) ) = ( 𝑃 ∨ ( 𝑁 ‘ 𝑃 ) ) )
17 10 15 12 16 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑃 ∨ ( 𝑅 ‘ 𝑁 ) ) = ( 𝑃 ∨ ( 𝑁 ‘ 𝑃 ) ) )
18 9 14 17 3eqtr3rd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ) ∧ ( ( 𝑅 ‘ 𝐹 ) = ( 𝑅 ‘ 𝑁 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ) ) → ( 𝑃 ∨ ( 𝑁 ‘ 𝑃 ) ) = ( ( 𝐹 ‘ 𝑃 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )