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Metamath Proof Explorer

Theorem posasymb

Description: A poset ordering is asymmetric. (Contributed by NM, 21-Oct-2011)

Ref Expression
Hypotheses posi.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
posi.l ⊢ ≤ = ( le ‘ 𝐾 )
Assertion posasymb ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) ↔ 𝑋 = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 posi.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 posi.l ⊢ ≤ = ( le ‘ 𝐾 )
3 simp1 ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝐾 ∈ Poset )
4 simp2 ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 )
5 simp3 ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑌 ∈ 𝐵 )
6 1 2 posi ⊢ ( ( 𝐾 ∈ Poset ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝑋 ≤ 𝑋 ∧ ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) → 𝑋 = 𝑌 ) ∧ ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑌 ) → 𝑋 ≤ 𝑌 ) ) )
7 3 4 5 5 6 syl13anc ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 ≤ 𝑋 ∧ ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) → 𝑋 = 𝑌 ) ∧ ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑌 ) → 𝑋 ≤ 𝑌 ) ) )
8 7 simp2d ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) → 𝑋 = 𝑌 ) )
9 1 2 posref ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ) → 𝑋 ≤ 𝑋 )
10 breq2 ⊢ ( 𝑋 = 𝑌 → ( 𝑋 ≤ 𝑋 ↔ 𝑋 ≤ 𝑌 ) )
11 9 10 syl5ibcom ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 = 𝑌 → 𝑋 ≤ 𝑌 ) )
12 breq1 ⊢ ( 𝑋 = 𝑌 → ( 𝑋 ≤ 𝑋 ↔ 𝑌 ≤ 𝑋 ) )
13 9 12 syl5ibcom ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 = 𝑌 → 𝑌 ≤ 𝑋 ) )
14 11 13 jcad ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 = 𝑌 → ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) ) )
15 14 3adant3 ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 = 𝑌 → ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) ) )
16 8 15 impbid ⊢ ( ( 𝐾 ∈ Poset ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 ≤ 𝑌 ∧ 𝑌 ≤ 𝑋 ) ↔ 𝑋 = 𝑌 ) )