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Metamath Proof Explorer

Theorem ltltncvr

Description: A chained strong ordering is not a covers relation. (Contributed by NM, 18-Jun-2012)

Ref Expression
Hypotheses ltltncvr.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
ltltncvr.s ⊢ < = ( lt ‘ 𝐾 )
ltltncvr.c ⊢ 𝐶 = ( ⋖ ‘ 𝐾 )
Assertion ltltncvr ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 < 𝑌 ∧ 𝑌 < 𝑍 ) → ¬ 𝑋 𝐶 𝑍 ) )

Proof

Step Hyp Ref Expression
1 ltltncvr.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 ltltncvr.s ⊢ < = ( lt ‘ 𝐾 )
3 ltltncvr.c ⊢ 𝐶 = ( ⋖ ‘ 𝐾 )
4 simpll ⊢ ( ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) ∧ 𝑋 𝐶 𝑍 ) → 𝐾 ∈ 𝐴 )
5 simplr1 ⊢ ( ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) ∧ 𝑋 𝐶 𝑍 ) → 𝑋 ∈ 𝐵 )
6 simplr3 ⊢ ( ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) ∧ 𝑋 𝐶 𝑍 ) → 𝑍 ∈ 𝐵 )
7 simplr2 ⊢ ( ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) ∧ 𝑋 𝐶 𝑍 ) → 𝑌 ∈ 𝐵 )
8 simpr ⊢ ( ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) ∧ 𝑋 𝐶 𝑍 ) → 𝑋 𝐶 𝑍 )
9 1 2 3 cvrnbtwn ⊢ ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ 𝑋 𝐶 𝑍 ) → ¬ ( 𝑋 < 𝑌 ∧ 𝑌 < 𝑍 ) )
10 4 5 6 7 8 9 syl131anc ⊢ ( ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) ∧ 𝑋 𝐶 𝑍 ) → ¬ ( 𝑋 < 𝑌 ∧ 𝑌 < 𝑍 ) )
11 10 ex ⊢ ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 𝐶 𝑍 → ¬ ( 𝑋 < 𝑌 ∧ 𝑌 < 𝑍 ) ) )
12 11 con2d ⊢ ( ( 𝐾 ∈ 𝐴 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 < 𝑌 ∧ 𝑌 < 𝑍 ) → ¬ 𝑋 𝐶 𝑍 ) )