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Metamath Proof Explorer

Theorem oppciso

Description: An isomorphism in the opposite category. See also remark 3.9 in Adamek p. 28. (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypotheses oppcsect.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
oppcsect.o ⊢ 𝑂 = ( oppCat ‘ 𝐶 )
oppcsect.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
oppcsect.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
oppcsect.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
oppciso.s ⊢ 𝐼 = ( Iso ‘ 𝐶 )
oppciso.t ⊢ 𝐽 = ( Iso ‘ 𝑂 )
Assertion oppciso ( 𝜑 → ( 𝑋 𝐽 𝑌 ) = ( 𝑌 𝐼 𝑋 ) )

Proof

Step Hyp Ref Expression
1 oppcsect.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
2 oppcsect.o ⊢ 𝑂 = ( oppCat ‘ 𝐶 )
3 oppcsect.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
4 oppcsect.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 oppcsect.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
6 oppciso.s ⊢ 𝐼 = ( Iso ‘ 𝐶 )
7 oppciso.t ⊢ 𝐽 = ( Iso ‘ 𝑂 )
8 eqid ⊢ ( Inv ‘ 𝐶 ) = ( Inv ‘ 𝐶 )
9 eqid ⊢ ( Inv ‘ 𝑂 ) = ( Inv ‘ 𝑂 )
10 1 2 3 4 5 8 9 oppcinv ⊢ ( 𝜑 → ( 𝑋 ( Inv ‘ 𝑂 ) 𝑌 ) = ( 𝑌 ( Inv ‘ 𝐶 ) 𝑋 ) )
11 10 dmeqd ⊢ ( 𝜑 → dom ( 𝑋 ( Inv ‘ 𝑂 ) 𝑌 ) = dom ( 𝑌 ( Inv ‘ 𝐶 ) 𝑋 ) )
12 2 1 oppcbas ⊢ 𝐵 = ( Base ‘ 𝑂 )
13 2 oppccat ⊢ ( 𝐶 ∈ Cat → 𝑂 ∈ Cat )
14 3 13 syl ⊢ ( 𝜑 → 𝑂 ∈ Cat )
15 12 9 14 4 5 7 isoval ⊢ ( 𝜑 → ( 𝑋 𝐽 𝑌 ) = dom ( 𝑋 ( Inv ‘ 𝑂 ) 𝑌 ) )
16 1 8 3 5 4 6 isoval ⊢ ( 𝜑 → ( 𝑌 𝐼 𝑋 ) = dom ( 𝑌 ( Inv ‘ 𝐶 ) 𝑋 ) )
17 11 15 16 3eqtr4d ⊢ ( 𝜑 → ( 𝑋 𝐽 𝑌 ) = ( 𝑌 𝐼 𝑋 ) )