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Metamath Proof Explorer

Theorem opcon3b

Description: Contraposition law for orthoposets. ( chcon3i analog.) (Contributed by NM, 8-Nov-2011)

Ref Expression
Hypotheses opoccl.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
opoccl.o ⊢ ⊥ = ( oc ‘ 𝐾 )
Assertion opcon3b ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 = 𝑌 ↔ ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 opoccl.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 opoccl.o ⊢ ⊥ = ( oc ‘ 𝐾 )
3 fveq2 ⊢ ( 𝑌 = 𝑋 → ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) )
4 3 eqcoms ⊢ ( 𝑋 = 𝑌 → ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) )
5 fveq2 ⊢ ( ( ⊥ ‘ 𝑋 ) = ( ⊥ ‘ 𝑌 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) )
6 5 eqcoms ⊢ ( ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) )
7 1 2 opococ ⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = 𝑋 )
8 7 3adant3 ⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = 𝑋 )
9 1 2 opococ ⊢ ( ( 𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) = 𝑌 )
10 9 3adant2 ⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) = 𝑌 )
11 8 10 eqeq12d ⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) ↔ 𝑋 = 𝑌 ) )
12 6 11 imbitrid ⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) → 𝑋 = 𝑌 ) )
13 4 12 impbid2 ⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 = 𝑌 ↔ ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) ) )