Metamath Proof Explorer


Theorem rlmsub

Description: Subtraction in the ring module. (Contributed by Thierry Arnoux, 30-Jun-2019)

Ref Expression
Assertion rlmsub ( -g ‘ 𝑅 ) = ( -g ‘ ( ringLMod ‘ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 rlmbas ⊢ ( Base ‘ 𝑅 ) = ( Base ‘ ( ringLMod ‘ 𝑅 ) )
2 1 a1i ⊢ ( ⊤ → ( Base ‘ 𝑅 ) = ( Base ‘ ( ringLMod ‘ 𝑅 ) ) )
3 rlmplusg ⊢ ( +g ‘ 𝑅 ) = ( +g ‘ ( ringLMod ‘ 𝑅 ) )
4 3 a1i ⊢ ( ⊤ → ( +g ‘ 𝑅 ) = ( +g ‘ ( ringLMod ‘ 𝑅 ) ) )
5 2 4 grpsubpropd ⊢ ( ⊤ → ( -g ‘ 𝑅 ) = ( -g ‘ ( ringLMod ‘ 𝑅 ) ) )
6 5 mptru ⊢ ( -g ‘ 𝑅 ) = ( -g ‘ ( ringLMod ‘ 𝑅 ) )