Metamath Proof Explorer


Theorem grpinvcld

Description: A group element's inverse is a group element. (Contributed by SN, 29-Jan-2025)

Ref Expression
Hypotheses grpinvcld.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpinvcld.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
grpinvcld.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
grpinvcld.1 ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
Assertion grpinvcld ( 𝜑 → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )

Proof

Step Hyp Ref Expression
1 grpinvcld.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpinvcld.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
3 grpinvcld.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
4 grpinvcld.1 ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 1 2 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )
6 3 4 5 syl2anc ⊢ ( 𝜑 → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )