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Metamath Proof Explorer

Theorem isline3

Description: Definition of line in terms of original lattice elements. (Contributed by NM, 29-Apr-2012)

Ref Expression
Hypotheses isline3.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
isline3.j ⊢ ∨ = ( join ‘ 𝐾 )
isline3.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
isline3.n ⊢ 𝑁 = ( Lines ‘ 𝐾 )
isline3.m ⊢ 𝑀 = ( pmap ‘ 𝐾 )
Assertion isline3 ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑀 ‘ 𝑋 ) ∈ 𝑁 ↔ ∃ 𝑝 ∈ 𝐴 ∃ 𝑞 ∈ 𝐴 ( 𝑝 ≠ 𝑞 ∧ 𝑋 = ( 𝑝 ∨ 𝑞 ) ) ) )

Proof

Step Hyp Ref Expression
1 isline3.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 isline3.j ⊢ ∨ = ( join ‘ 𝐾 )
3 isline3.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
4 isline3.n ⊢ 𝑁 = ( Lines ‘ 𝐾 )
5 isline3.m ⊢ 𝑀 = ( pmap ‘ 𝐾 )
6 hllat ⊢ ( 𝐾 ∈ HL → 𝐾 ∈ Lat )
7 6 adantr ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) → 𝐾 ∈ Lat )
8 2 3 4 5 isline2 ⊢ ( 𝐾 ∈ Lat → ( ( 𝑀 ‘ 𝑋 ) ∈ 𝑁 ↔ ∃ 𝑝 ∈ 𝐴 ∃ 𝑞 ∈ 𝐴 ( 𝑝 ≠ 𝑞 ∧ ( 𝑀 ‘ 𝑋 ) = ( 𝑀 ‘ ( 𝑝 ∨ 𝑞 ) ) ) ) )
9 7 8 syl ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑀 ‘ 𝑋 ) ∈ 𝑁 ↔ ∃ 𝑝 ∈ 𝐴 ∃ 𝑞 ∈ 𝐴 ( 𝑝 ≠ 𝑞 ∧ ( 𝑀 ‘ 𝑋 ) = ( 𝑀 ‘ ( 𝑝 ∨ 𝑞 ) ) ) ) )
10 simpll ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → 𝐾 ∈ HL )
11 simplr ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → 𝑋 ∈ 𝐵 )
12 6 ad2antrr ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → 𝐾 ∈ Lat )
13 1 3 atbase ⊢ ( 𝑝 ∈ 𝐴 → 𝑝 ∈ 𝐵 )
14 13 ad2antrl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → 𝑝 ∈ 𝐵 )
15 1 3 atbase ⊢ ( 𝑞 ∈ 𝐴 → 𝑞 ∈ 𝐵 )
16 15 ad2antll ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → 𝑞 ∈ 𝐵 )
17 1 2 latjcl ⊢ ( ( 𝐾 ∈ Lat ∧ 𝑝 ∈ 𝐵 ∧ 𝑞 ∈ 𝐵 ) → ( 𝑝 ∨ 𝑞 ) ∈ 𝐵 )
18 12 14 16 17 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → ( 𝑝 ∨ 𝑞 ) ∈ 𝐵 )
19 1 5 pmap11 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ∧ ( 𝑝 ∨ 𝑞 ) ∈ 𝐵 ) → ( ( 𝑀 ‘ 𝑋 ) = ( 𝑀 ‘ ( 𝑝 ∨ 𝑞 ) ) ↔ 𝑋 = ( 𝑝 ∨ 𝑞 ) ) )
20 10 11 18 19 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → ( ( 𝑀 ‘ 𝑋 ) = ( 𝑀 ‘ ( 𝑝 ∨ 𝑞 ) ) ↔ 𝑋 = ( 𝑝 ∨ 𝑞 ) ) )
21 20 anbi2d ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) ∧ ( 𝑝 ∈ 𝐴 ∧ 𝑞 ∈ 𝐴 ) ) → ( ( 𝑝 ≠ 𝑞 ∧ ( 𝑀 ‘ 𝑋 ) = ( 𝑀 ‘ ( 𝑝 ∨ 𝑞 ) ) ) ↔ ( 𝑝 ≠ 𝑞 ∧ 𝑋 = ( 𝑝 ∨ 𝑞 ) ) ) )
22 21 2rexbidva ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) → ( ∃ 𝑝 ∈ 𝐴 ∃ 𝑞 ∈ 𝐴 ( 𝑝 ≠ 𝑞 ∧ ( 𝑀 ‘ 𝑋 ) = ( 𝑀 ‘ ( 𝑝 ∨ 𝑞 ) ) ) ↔ ∃ 𝑝 ∈ 𝐴 ∃ 𝑞 ∈ 𝐴 ( 𝑝 ≠ 𝑞 ∧ 𝑋 = ( 𝑝 ∨ 𝑞 ) ) ) )
23 9 22 bitrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑀 ‘ 𝑋 ) ∈ 𝑁 ↔ ∃ 𝑝 ∈ 𝐴 ∃ 𝑞 ∈ 𝐴 ( 𝑝 ≠ 𝑞 ∧ 𝑋 = ( 𝑝 ∨ 𝑞 ) ) ) )