Metamath Proof Explorer


Theorem cdleme46frvlpq

Description: Show that ( FS ) is not under P .\/ Q when S isn't. (Contributed by NM, 1-Apr-2013)

Ref Expression
Hypotheses cdlemef46.b 𝐵 = ( Base ‘ 𝐾 )
cdlemef46.l = ( le ‘ 𝐾 )
cdlemef46.j = ( join ‘ 𝐾 )
cdlemef46.m = ( meet ‘ 𝐾 )
cdlemef46.a 𝐴 = ( Atoms ‘ 𝐾 )
cdlemef46.h 𝐻 = ( LHyp ‘ 𝐾 )
cdlemef46.u 𝑈 = ( ( 𝑃 𝑄 ) 𝑊 )
cdlemef46.d 𝐷 = ( ( 𝑡 𝑈 ) ( 𝑄 ( ( 𝑃 𝑡 ) 𝑊 ) ) )
cdlemefs46.e 𝐸 = ( ( 𝑃 𝑄 ) ( 𝐷 ( ( 𝑠 𝑡 ) 𝑊 ) ) )
cdlemef46.f 𝐹 = ( 𝑥𝐵 ↦ if ( ( 𝑃𝑄 ∧ ¬ 𝑥 𝑊 ) , ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑥 𝑊 ) ) = 𝑥 ) → 𝑧 = ( if ( 𝑠 ( 𝑃 𝑄 ) , ( 𝑦𝐵𝑡𝐴 ( ( ¬ 𝑡 𝑊 ∧ ¬ 𝑡 ( 𝑃 𝑄 ) ) → 𝑦 = 𝐸 ) ) , 𝑠 / 𝑡 𝐷 ) ( 𝑥 𝑊 ) ) ) ) , 𝑥 ) )
Assertion cdleme46frvlpq ( ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ ( 𝑄𝐴 ∧ ¬ 𝑄 𝑊 ) ) ∧ ( 𝑃𝑄 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ¬ ( 𝐹𝑆 ) ( 𝑃 𝑄 ) )

Proof

Step Hyp Ref Expression
1 cdlemef46.b 𝐵 = ( Base ‘ 𝐾 )
2 cdlemef46.l = ( le ‘ 𝐾 )
3 cdlemef46.j = ( join ‘ 𝐾 )
4 cdlemef46.m = ( meet ‘ 𝐾 )
5 cdlemef46.a 𝐴 = ( Atoms ‘ 𝐾 )
6 cdlemef46.h 𝐻 = ( LHyp ‘ 𝐾 )
7 cdlemef46.u 𝑈 = ( ( 𝑃 𝑄 ) 𝑊 )
8 cdlemef46.d 𝐷 = ( ( 𝑡 𝑈 ) ( 𝑄 ( ( 𝑃 𝑡 ) 𝑊 ) ) )
9 cdlemefs46.e 𝐸 = ( ( 𝑃 𝑄 ) ( 𝐷 ( ( 𝑠 𝑡 ) 𝑊 ) ) )
10 cdlemef46.f 𝐹 = ( 𝑥𝐵 ↦ if ( ( 𝑃𝑄 ∧ ¬ 𝑥 𝑊 ) , ( 𝑧𝐵𝑠𝐴 ( ( ¬ 𝑠 𝑊 ∧ ( 𝑠 ( 𝑥 𝑊 ) ) = 𝑥 ) → 𝑧 = ( if ( 𝑠 ( 𝑃 𝑄 ) , ( 𝑦𝐵𝑡𝐴 ( ( ¬ 𝑡 𝑊 ∧ ¬ 𝑡 ( 𝑃 𝑄 ) ) → 𝑦 = 𝐸 ) ) , 𝑠 / 𝑡 𝐷 ) ( 𝑥 𝑊 ) ) ) ) , 𝑥 ) )
11 eqid ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) = ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) )
12 2 3 4 5 6 7 11 cdleme35fnpq ( ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ ( 𝑄𝐴 ∧ ¬ 𝑄 𝑊 ) ) ∧ ( 𝑃𝑄 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ¬ ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) ( 𝑃 𝑄 ) )
13 1 2 3 4 5 6 7 8 10 cdlemefr45e ( ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ ( 𝑄𝐴 ∧ ¬ 𝑄 𝑊 ) ) ∧ ( 𝑃𝑄 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ( 𝐹𝑆 ) = ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) )
14 13 breq1d ( ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ ( 𝑄𝐴 ∧ ¬ 𝑄 𝑊 ) ) ∧ ( 𝑃𝑄 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ( ( 𝐹𝑆 ) ( 𝑃 𝑄 ) ↔ ( ( 𝑆 𝑈 ) ( 𝑄 ( ( 𝑃 𝑆 ) 𝑊 ) ) ) ( 𝑃 𝑄 ) ) )
15 12 14 mtbird ( ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ∧ ( 𝑄𝐴 ∧ ¬ 𝑄 𝑊 ) ) ∧ ( 𝑃𝑄 ∧ ( 𝑆𝐴 ∧ ¬ 𝑆 𝑊 ) ) ∧ ¬ 𝑆 ( 𝑃 𝑄 ) ) → ¬ ( 𝐹𝑆 ) ( 𝑃 𝑄 ) )