Metamath Proof Explorer


Theorem osumcllem4N

Description: Lemma for osumclN . (Contributed by NM, 24-Mar-2012) (New usage is discouraged.)

Ref Expression
Hypotheses osumcllem.l = ( le ‘ 𝐾 )
osumcllem.j = ( join ‘ 𝐾 )
osumcllem.a 𝐴 = ( Atoms ‘ 𝐾 )
osumcllem.p + = ( +𝑃𝐾 )
osumcllem.o = ( ⊥𝑃𝐾 )
osumcllem.c 𝐶 = ( PSubCl ‘ 𝐾 )
osumcllem.m 𝑀 = ( 𝑋 + { 𝑝 } )
osumcllem.u 𝑈 = ( ‘ ( ‘ ( 𝑋 + 𝑌 ) ) )
Assertion osumcllem4N ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → 𝑞𝑟 )

Proof

Step Hyp Ref Expression
1 osumcllem.l = ( le ‘ 𝐾 )
2 osumcllem.j = ( join ‘ 𝐾 )
3 osumcllem.a 𝐴 = ( Atoms ‘ 𝐾 )
4 osumcllem.p + = ( +𝑃𝐾 )
5 osumcllem.o = ( ⊥𝑃𝐾 )
6 osumcllem.c 𝐶 = ( PSubCl ‘ 𝐾 )
7 osumcllem.m 𝑀 = ( 𝑋 + { 𝑝 } )
8 osumcllem.u 𝑈 = ( ‘ ( ‘ ( 𝑋 + 𝑌 ) ) )
9 n0i ( 𝑟 ∈ ( 𝑋𝑌 ) → ¬ ( 𝑋𝑌 ) = ∅ )
10 incom ( 𝑋𝑌 ) = ( 𝑌𝑋 )
11 sslin ( 𝑋 ⊆ ( 𝑌 ) → ( 𝑌𝑋 ) ⊆ ( 𝑌 ∩ ( 𝑌 ) ) )
12 11 3ad2ant3 ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) → ( 𝑌𝑋 ) ⊆ ( 𝑌 ∩ ( 𝑌 ) ) )
13 10 12 eqsstrid ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) → ( 𝑋𝑌 ) ⊆ ( 𝑌 ∩ ( 𝑌 ) ) )
14 3 5 pnonsingN ( ( 𝐾 ∈ HL ∧ 𝑌𝐴 ) → ( 𝑌 ∩ ( 𝑌 ) ) = ∅ )
15 14 3adant3 ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) → ( 𝑌 ∩ ( 𝑌 ) ) = ∅ )
16 13 15 sseqtrd ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) → ( 𝑋𝑌 ) ⊆ ∅ )
17 ss0b ( ( 𝑋𝑌 ) ⊆ ∅ ↔ ( 𝑋𝑌 ) = ∅ )
18 16 17 sylib ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) → ( 𝑋𝑌 ) = ∅ )
19 18 adantr ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → ( 𝑋𝑌 ) = ∅ )
20 9 19 nsyl3 ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → ¬ 𝑟 ∈ ( 𝑋𝑌 ) )
21 simprr ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → 𝑞𝑌 )
22 eleq1w ( 𝑞 = 𝑟 → ( 𝑞𝑌𝑟𝑌 ) )
23 21 22 syl5ibcom ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → ( 𝑞 = 𝑟𝑟𝑌 ) )
24 simprl ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → 𝑟𝑋 )
25 23 24 jctild ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → ( 𝑞 = 𝑟 → ( 𝑟𝑋𝑟𝑌 ) ) )
26 elin ( 𝑟 ∈ ( 𝑋𝑌 ) ↔ ( 𝑟𝑋𝑟𝑌 ) )
27 25 26 syl6ibr ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → ( 𝑞 = 𝑟𝑟 ∈ ( 𝑋𝑌 ) ) )
28 27 necon3bd ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → ( ¬ 𝑟 ∈ ( 𝑋𝑌 ) → 𝑞𝑟 ) )
29 20 28 mpd ( ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) ∧ ( 𝑟𝑋𝑞𝑌 ) ) → 𝑞𝑟 )