Metamath Proof Explorer


Theorem omndmnd

Description: A left-ordered monoid is a monoid. (Contributed by Thierry Arnoux, 13-Mar-2018)

Ref Expression
Assertion omndmnd ( 𝑀 ∈ oMnd → 𝑀 ∈ Mnd )

Proof

Step Hyp Ref Expression
1 eqid ( Base ‘ 𝑀 ) = ( Base ‘ 𝑀 )
2 eqid ( +g𝑀 ) = ( +g𝑀 )
3 eqid ( le ‘ 𝑀 ) = ( le ‘ 𝑀 )
4 1 2 3 isomnd ( 𝑀 ∈ oMnd ↔ ( 𝑀 ∈ Mnd ∧ 𝑀 ∈ Toset ∧ ∀ 𝑎 ∈ ( Base ‘ 𝑀 ) ∀ 𝑏 ∈ ( Base ‘ 𝑀 ) ∀ 𝑐 ∈ ( Base ‘ 𝑀 ) ( 𝑎 ( le ‘ 𝑀 ) 𝑏 → ( 𝑎 ( +g𝑀 ) 𝑐 ) ( le ‘ 𝑀 ) ( 𝑏 ( +g𝑀 ) 𝑐 ) ) ) )
5 4 simp1bi ( 𝑀 ∈ oMnd → 𝑀 ∈ Mnd )