Metamath Proof Explorer


Theorem 2arwcatlem2

Description: Lemma for 2arwcat . (Contributed by Zhi Wang, 5-Nov-2025)

Ref Expression
Hypotheses 2arwcatlem2.a ( 𝜑𝐴 = 𝑋 )
2arwcatlem2.b ( 𝜑𝐵 = 𝑌 )
2arwcatlem2.c ( 𝜑𝐶 = 𝑍 )
2arwcatlem2.f ( 𝜑 → ( 𝐹 = 0𝐹 = 1 ) )
2arwcatlem2.1 ( 𝜑 → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 1 ) = 1 )
2arwcatlem2.0 ( 𝜑 → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 0 ) = 0 )
Assertion 2arwcatlem2 ( 𝜑 → ( 1 ( ⟨ 𝐴 , 𝐵· 𝐶 ) 𝐹 ) = 𝐹 )

Proof

Step Hyp Ref Expression
1 2arwcatlem2.a ( 𝜑𝐴 = 𝑋 )
2 2arwcatlem2.b ( 𝜑𝐵 = 𝑌 )
3 2arwcatlem2.c ( 𝜑𝐶 = 𝑍 )
4 2arwcatlem2.f ( 𝜑 → ( 𝐹 = 0𝐹 = 1 ) )
5 2arwcatlem2.1 ( 𝜑 → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 1 ) = 1 )
6 2arwcatlem2.0 ( 𝜑 → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 0 ) = 0 )
7 1 2 opeq12d ( 𝜑 → ⟨ 𝐴 , 𝐵 ⟩ = ⟨ 𝑋 , 𝑌 ⟩ )
8 7 3 oveq12d ( 𝜑 → ( ⟨ 𝐴 , 𝐵· 𝐶 ) = ( ⟨ 𝑋 , 𝑌· 𝑍 ) )
9 8 oveqd ( 𝜑 → ( 1 ( ⟨ 𝐴 , 𝐵· 𝐶 ) 𝐹 ) = ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 𝐹 ) )
10 6 adantr ( ( 𝜑𝐹 = 0 ) → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 0 ) = 0 )
11 simpr ( ( 𝜑𝐹 = 0 ) → 𝐹 = 0 )
12 11 oveq2d ( ( 𝜑𝐹 = 0 ) → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 𝐹 ) = ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 0 ) )
13 10 12 11 3eqtr4d ( ( 𝜑𝐹 = 0 ) → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 𝐹 ) = 𝐹 )
14 5 adantr ( ( 𝜑𝐹 = 1 ) → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 1 ) = 1 )
15 simpr ( ( 𝜑𝐹 = 1 ) → 𝐹 = 1 )
16 15 oveq2d ( ( 𝜑𝐹 = 1 ) → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 𝐹 ) = ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 1 ) )
17 14 16 15 3eqtr4d ( ( 𝜑𝐹 = 1 ) → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 𝐹 ) = 𝐹 )
18 13 17 4 mpjaodan ( 𝜑 → ( 1 ( ⟨ 𝑋 , 𝑌· 𝑍 ) 𝐹 ) = 𝐹 )
19 9 18 eqtrd ( 𝜑 → ( 1 ( ⟨ 𝐴 , 𝐵· 𝐶 ) 𝐹 ) = 𝐹 )