Metamath Proof Explorer


Theorem ncvr1

Description: No element covers the lattice unit. (Contributed by NM, 8-Jul-2013)

Ref Expression
Hypotheses ncvr1.b 𝐵 = ( Base ‘ 𝐾 )
ncvr1.u 1 = ( 1. ‘ 𝐾 )
ncvr1.c 𝐶 = ( ⋖ ‘ 𝐾 )
Assertion ncvr1 ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) → ¬ 1 𝐶 𝑋 )

Proof

Step Hyp Ref Expression
1 ncvr1.b 𝐵 = ( Base ‘ 𝐾 )
2 ncvr1.u 1 = ( 1. ‘ 𝐾 )
3 ncvr1.c 𝐶 = ( ⋖ ‘ 𝐾 )
4 eqid ( le ‘ 𝐾 ) = ( le ‘ 𝐾 )
5 1 4 2 ople1 ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) → 𝑋 ( le ‘ 𝐾 ) 1 )
6 opposet ( 𝐾 ∈ OP → 𝐾 ∈ Poset )
7 6 ad2antrr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 ( lt ‘ 𝐾 ) 𝑋 ) → 𝐾 ∈ Poset )
8 1 2 op1cl ( 𝐾 ∈ OP → 1𝐵 )
9 8 ad2antrr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 ( lt ‘ 𝐾 ) 𝑋 ) → 1𝐵 )
10 simplr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 ( lt ‘ 𝐾 ) 𝑋 ) → 𝑋𝐵 )
11 simpr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 ( lt ‘ 𝐾 ) 𝑋 ) → 1 ( lt ‘ 𝐾 ) 𝑋 )
12 eqid ( lt ‘ 𝐾 ) = ( lt ‘ 𝐾 )
13 1 4 12 pltnle ( ( ( 𝐾 ∈ Poset ∧ 1𝐵𝑋𝐵 ) ∧ 1 ( lt ‘ 𝐾 ) 𝑋 ) → ¬ 𝑋 ( le ‘ 𝐾 ) 1 )
14 7 9 10 11 13 syl31anc ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 ( lt ‘ 𝐾 ) 𝑋 ) → ¬ 𝑋 ( le ‘ 𝐾 ) 1 )
15 14 ex ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) → ( 1 ( lt ‘ 𝐾 ) 𝑋 → ¬ 𝑋 ( le ‘ 𝐾 ) 1 ) )
16 5 15 mt2d ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) → ¬ 1 ( lt ‘ 𝐾 ) 𝑋 )
17 simpll ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 𝐶 𝑋 ) → 𝐾 ∈ OP )
18 8 ad2antrr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 𝐶 𝑋 ) → 1𝐵 )
19 simplr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 𝐶 𝑋 ) → 𝑋𝐵 )
20 simpr ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 𝐶 𝑋 ) → 1 𝐶 𝑋 )
21 1 12 3 cvrlt ( ( ( 𝐾 ∈ OP ∧ 1𝐵𝑋𝐵 ) ∧ 1 𝐶 𝑋 ) → 1 ( lt ‘ 𝐾 ) 𝑋 )
22 17 18 19 20 21 syl31anc ( ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) ∧ 1 𝐶 𝑋 ) → 1 ( lt ‘ 𝐾 ) 𝑋 )
23 16 22 mtand ( ( 𝐾 ∈ OP ∧ 𝑋𝐵 ) → ¬ 1 𝐶 𝑋 )