Metamath Proof Explorer


Theorem oppgoppcco

Description: The converted opposite monoid has the same composition as that of the opposite category. Example 3.6(2) of Adamek p. 25. (Contributed by Zhi Wang, 22-Sep-2025)

Ref Expression
Hypotheses mndtccat.c ( 𝜑𝐶 = ( MndToCat ‘ 𝑀 ) )
mndtccat.m ( 𝜑𝑀 ∈ Mnd )
oppgoppchom.d ( 𝜑𝐷 = ( MndToCat ‘ ( oppg𝑀 ) ) )
oppgoppchom.o 𝑂 = ( oppCat ‘ 𝐶 )
oppgoppchom.x ( 𝜑𝑋 ∈ ( Base ‘ 𝐷 ) )
oppgoppchom.y ( 𝜑𝑌 ∈ ( Base ‘ 𝑂 ) )
oppgoppcco.o ( 𝜑· = ( comp ‘ 𝐷 ) )
oppgoppcco.x ( 𝜑 = ( comp ‘ 𝑂 ) )
Assertion oppgoppcco ( 𝜑 → ( ⟨ 𝑋 , 𝑋· 𝑋 ) = ( ⟨ 𝑌 , 𝑌 𝑌 ) )

Proof

Step Hyp Ref Expression
1 mndtccat.c ( 𝜑𝐶 = ( MndToCat ‘ 𝑀 ) )
2 mndtccat.m ( 𝜑𝑀 ∈ Mnd )
3 oppgoppchom.d ( 𝜑𝐷 = ( MndToCat ‘ ( oppg𝑀 ) ) )
4 oppgoppchom.o 𝑂 = ( oppCat ‘ 𝐶 )
5 oppgoppchom.x ( 𝜑𝑋 ∈ ( Base ‘ 𝐷 ) )
6 oppgoppchom.y ( 𝜑𝑌 ∈ ( Base ‘ 𝑂 ) )
7 oppgoppcco.o ( 𝜑· = ( comp ‘ 𝐷 ) )
8 oppgoppcco.x ( 𝜑 = ( comp ‘ 𝑂 ) )
9 eqid ( Base ‘ 𝐶 ) = ( Base ‘ 𝐶 )
10 4 9 oppcbas ( Base ‘ 𝐶 ) = ( Base ‘ 𝑂 )
11 10 eqcomi ( Base ‘ 𝑂 ) = ( Base ‘ 𝐶 )
12 11 a1i ( 𝜑 → ( Base ‘ 𝑂 ) = ( Base ‘ 𝐶 ) )
13 eqidd ( 𝜑 → ( comp ‘ 𝐶 ) = ( comp ‘ 𝐶 ) )
14 1 2 12 6 6 6 13 mndtcco ( 𝜑 → ( ⟨ 𝑌 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑌 ) = ( +g𝑀 ) )
15 14 tposeqd ( 𝜑 → tpos ( ⟨ 𝑌 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑌 ) = tpos ( +g𝑀 ) )
16 eqid ( comp ‘ 𝐶 ) = ( comp ‘ 𝐶 )
17 11 16 4 6 6 6 oppccofval ( 𝜑 → ( ⟨ 𝑌 , 𝑌 ⟩ ( comp ‘ 𝑂 ) 𝑌 ) = tpos ( ⟨ 𝑌 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑌 ) )
18 eqid ( oppg𝑀 ) = ( oppg𝑀 )
19 18 oppgmnd ( 𝑀 ∈ Mnd → ( oppg𝑀 ) ∈ Mnd )
20 2 19 syl ( 𝜑 → ( oppg𝑀 ) ∈ Mnd )
21 eqidd ( 𝜑 → ( Base ‘ 𝐷 ) = ( Base ‘ 𝐷 ) )
22 3 20 21 5 5 5 7 mndtcco ( 𝜑 → ( ⟨ 𝑋 , 𝑋· 𝑋 ) = ( +g ‘ ( oppg𝑀 ) ) )
23 eqid ( +g𝑀 ) = ( +g𝑀 )
24 eqid ( +g ‘ ( oppg𝑀 ) ) = ( +g ‘ ( oppg𝑀 ) )
25 23 18 24 oppgplusfval ( +g ‘ ( oppg𝑀 ) ) = tpos ( +g𝑀 )
26 22 25 eqtrdi ( 𝜑 → ( ⟨ 𝑋 , 𝑋· 𝑋 ) = tpos ( +g𝑀 ) )
27 15 17 26 3eqtr4rd ( 𝜑 → ( ⟨ 𝑋 , 𝑋· 𝑋 ) = ( ⟨ 𝑌 , 𝑌 ⟩ ( comp ‘ 𝑂 ) 𝑌 ) )
28 8 oveqd ( 𝜑 → ( ⟨ 𝑌 , 𝑌 𝑌 ) = ( ⟨ 𝑌 , 𝑌 ⟩ ( comp ‘ 𝑂 ) 𝑌 ) )
29 27 28 eqtr4d ( 𝜑 → ( ⟨ 𝑋 , 𝑋· 𝑋 ) = ( ⟨ 𝑌 , 𝑌 𝑌 ) )