Metamath Proof Explorer


Theorem atcvrj0

Description: Two atoms covering the zero subspace are equal. ( atcv1 analog.) (Contributed by NM, 29-Nov-2011)

Ref Expression
Hypotheses atcvrj0.b 𝐵 = ( Base ‘ 𝐾 )
atcvrj0.j = ( join ‘ 𝐾 )
atcvrj0.z 0 = ( 0. ‘ 𝐾 )
atcvrj0.c 𝐶 = ( ⋖ ‘ 𝐾 )
atcvrj0.a 𝐴 = ( Atoms ‘ 𝐾 )
Assertion atcvrj0 ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ∧ 𝑋 𝐶 ( 𝑃 𝑄 ) ) → ( 𝑋 = 0𝑃 = 𝑄 ) )

Proof

Step Hyp Ref Expression
1 atcvrj0.b 𝐵 = ( Base ‘ 𝐾 )
2 atcvrj0.j = ( join ‘ 𝐾 )
3 atcvrj0.z 0 = ( 0. ‘ 𝐾 )
4 atcvrj0.c 𝐶 = ( ⋖ ‘ 𝐾 )
5 atcvrj0.a 𝐴 = ( Atoms ‘ 𝐾 )
6 breq1 ( 𝑋 = 0 → ( 𝑋 𝐶 ( 𝑃 𝑄 ) ↔ 0 𝐶 ( 𝑃 𝑄 ) ) )
7 6 biimpd ( 𝑋 = 0 → ( 𝑋 𝐶 ( 𝑃 𝑄 ) → 0 𝐶 ( 𝑃 𝑄 ) ) )
8 7 adantl ( ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) ∧ 𝑋 = 0 ) → ( 𝑋 𝐶 ( 𝑃 𝑄 ) → 0 𝐶 ( 𝑃 𝑄 ) ) )
9 2 3 4 5 atcvr0eq ( ( 𝐾 ∈ HL ∧ 𝑃𝐴𝑄𝐴 ) → ( 0 𝐶 ( 𝑃 𝑄 ) ↔ 𝑃 = 𝑄 ) )
10 9 3adant3r1 ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 0 𝐶 ( 𝑃 𝑄 ) ↔ 𝑃 = 𝑄 ) )
11 10 adantr ( ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) ∧ 𝑋 = 0 ) → ( 0 𝐶 ( 𝑃 𝑄 ) ↔ 𝑃 = 𝑄 ) )
12 8 11 sylibd ( ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) ∧ 𝑋 = 0 ) → ( 𝑋 𝐶 ( 𝑃 𝑄 ) → 𝑃 = 𝑄 ) )
13 12 ex ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 = 0 → ( 𝑋 𝐶 ( 𝑃 𝑄 ) → 𝑃 = 𝑄 ) ) )
14 13 com23 ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 𝐶 ( 𝑃 𝑄 ) → ( 𝑋 = 0𝑃 = 𝑄 ) ) )
15 14 3impia ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ∧ 𝑋 𝐶 ( 𝑃 𝑄 ) ) → ( 𝑋 = 0𝑃 = 𝑄 ) )
16 oveq1 ( 𝑃 = 𝑄 → ( 𝑃 𝑄 ) = ( 𝑄 𝑄 ) )
17 16 breq2d ( 𝑃 = 𝑄 → ( 𝑋 𝐶 ( 𝑃 𝑄 ) ↔ 𝑋 𝐶 ( 𝑄 𝑄 ) ) )
18 17 biimpac ( ( 𝑋 𝐶 ( 𝑃 𝑄 ) ∧ 𝑃 = 𝑄 ) → 𝑋 𝐶 ( 𝑄 𝑄 ) )
19 simpr3 ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → 𝑄𝐴 )
20 2 5 hlatjidm ( ( 𝐾 ∈ HL ∧ 𝑄𝐴 ) → ( 𝑄 𝑄 ) = 𝑄 )
21 19 20 syldan ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑄 𝑄 ) = 𝑄 )
22 21 breq2d ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 𝐶 ( 𝑄 𝑄 ) ↔ 𝑋 𝐶 𝑄 ) )
23 hlatl ( 𝐾 ∈ HL → 𝐾 ∈ AtLat )
24 23 adantr ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → 𝐾 ∈ AtLat )
25 simpr1 ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → 𝑋𝐵 )
26 eqid ( le ‘ 𝐾 ) = ( le ‘ 𝐾 )
27 1 26 3 4 5 atcvreq0 ( ( 𝐾 ∈ AtLat ∧ 𝑋𝐵𝑄𝐴 ) → ( 𝑋 𝐶 𝑄𝑋 = 0 ) )
28 24 25 19 27 syl3anc ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 𝐶 𝑄𝑋 = 0 ) )
29 28 biimpd ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 𝐶 𝑄𝑋 = 0 ) )
30 22 29 sylbid ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 𝐶 ( 𝑄 𝑄 ) → 𝑋 = 0 ) )
31 18 30 syl5 ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( ( 𝑋 𝐶 ( 𝑃 𝑄 ) ∧ 𝑃 = 𝑄 ) → 𝑋 = 0 ) )
32 31 expd ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ) → ( 𝑋 𝐶 ( 𝑃 𝑄 ) → ( 𝑃 = 𝑄𝑋 = 0 ) ) )
33 32 3impia ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ∧ 𝑋 𝐶 ( 𝑃 𝑄 ) ) → ( 𝑃 = 𝑄𝑋 = 0 ) )
34 15 33 impbid ( ( 𝐾 ∈ HL ∧ ( 𝑋𝐵𝑃𝐴𝑄𝐴 ) ∧ 𝑋 𝐶 ( 𝑃 𝑄 ) ) → ( 𝑋 = 0𝑃 = 𝑄 ) )