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Metamath Proof Explorer

Theorem grpsubinv

Description: Subtraction of an inverse. (Contributed by NM, 7-Apr-2015)

Ref Expression
Hypotheses grpsubinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpsubinv.p ⊢ + = ( +g ‘ 𝐺 )
grpsubinv.m ⊢ − = ( -g ‘ 𝐺 )
grpsubinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
grpsubinv.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
grpsubinv.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
grpsubinv.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
Assertion grpsubinv ( 𝜑 → ( 𝑋 − ( 𝑁 ‘ 𝑌 ) ) = ( 𝑋 + 𝑌 ) )

Proof

Step Hyp Ref Expression
1 grpsubinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpsubinv.p ⊢ + = ( +g ‘ 𝐺 )
3 grpsubinv.m ⊢ − = ( -g ‘ 𝐺 )
4 grpsubinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
5 grpsubinv.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
6 grpsubinv.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
7 grpsubinv.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
8 1 4 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 )
9 5 7 8 syl2anc ⊢ ( 𝜑 → ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 )
10 1 2 4 3 grpsubval ⊢ ( ( 𝑋 ∈ 𝐵 ∧ ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 ) → ( 𝑋 − ( 𝑁 ‘ 𝑌 ) ) = ( 𝑋 + ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) ) )
11 6 9 10 syl2anc ⊢ ( 𝜑 → ( 𝑋 − ( 𝑁 ‘ 𝑌 ) ) = ( 𝑋 + ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) ) )
12 1 4 grpinvinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) = 𝑌 )
13 5 7 12 syl2anc ⊢ ( 𝜑 → ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) = 𝑌 )
14 13 oveq2d ⊢ ( 𝜑 → ( 𝑋 + ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) ) = ( 𝑋 + 𝑌 ) )
15 11 14 eqtrd ⊢ ( 𝜑 → ( 𝑋 − ( 𝑁 ‘ 𝑌 ) ) = ( 𝑋 + 𝑌 ) )