Metamath Proof Explorer


Theorem cdleme17d1

Description: Part of proof of Lemma E in Crawley p. 114, first part of 4th paragraph. F , G represent f(s), f_s(p) respectively. We show, in their notation, f_s(p)=q. (Contributed by NM, 11-Oct-2012)

Ref Expression
Hypotheses cdleme17.l = ( le ‘ 𝐾 )
cdleme17.j = ( join ‘ 𝐾 )
cdleme17.m = ( meet ‘ 𝐾 )
cdleme17.a 𝐴 = ( Atoms ‘ 𝐾 )
cdleme17.h 𝐻 = ( LHyp ‘ 𝐾 )
cdleme17.u 𝑈 = ( ( 𝑃 𝑄 ) 𝑊 )
cdleme17.f 𝐹 = ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) )
cdleme17.g 𝐺 = ( ( 𝑃 𝑄 ) ( 𝐹 ( ( 𝑃 𝑆 ) 𝑊 ) ) )
Assertion cdleme17d1 ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝐺 = 𝑄 )

Proof

Step Hyp Ref Expression
1 cdleme17.l = ( le ‘ 𝐾 )
2 cdleme17.j = ( join ‘ 𝐾 )
3 cdleme17.m = ( meet ‘ 𝐾 )
4 cdleme17.a 𝐴 = ( Atoms ‘ 𝐾 )
5 cdleme17.h 𝐻 = ( LHyp ‘ 𝐾 )
6 cdleme17.u 𝑈 = ( ( 𝑃 𝑄 ) 𝑊 )
7 cdleme17.f 𝐹 = ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) )
8 cdleme17.g 𝐺 = ( ( 𝑃 𝑄 ) ( 𝐹 ( ( 𝑃 𝑆 ) 𝑊 ) ) )
9 eqid ( ( 𝑃 𝑆 ) 𝑊 ) = ( ( 𝑃 𝑆 ) 𝑊 )
10 1 2 3 4 5 6 7 8 9 cdleme17a ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝐺 = ( ( 𝑃 𝑄 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) )
11 simp1l ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝐾 ∈ HL )
12 simp1r ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝑊𝐻 )
13 simp21l ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝑃𝐴 )
14 simp21r ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ¬ 𝑃 𝑊 )
15 simp22 ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝑄𝐴 )
16 simp23l ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝑆𝐴 )
17 simp3 ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ¬ 𝑆 ( 𝑃 𝑄 ) )
18 1 2 3 4 5 6 7 8 9 cdleme17c ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ ( 𝑄𝐴𝑆𝐴 ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) ) → ( ( 𝑃 𝑄 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) = 𝑄 )
19 11 12 13 14 15 16 17 18 syl223anc ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ( ( 𝑃 𝑄 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) = 𝑄 )
20 10 19 eqtrd ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ 𝑄𝐴 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → 𝐺 = 𝑄 )