Metamath Proof Explorer

Theorem grpnpcan

Description: Cancellation law for subtraction ( npcan analog). (Contributed by NM, 19-Apr-2014)

Ref Expression
Hypotheses grpsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpsubadd.p ⊢ + = ( +g ‘ 𝐺 )
grpsubadd.m ⊢ − = ( -g ‘ 𝐺 )
Assertion grpnpcan ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 − 𝑌 ) + 𝑌 ) = 𝑋 )

Proof

Step Hyp Ref Expression
1 grpsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpsubadd.p ⊢ + = ( +g ‘ 𝐺 )
3 grpsubadd.m ⊢ − = ( -g ‘ 𝐺 )
4 eqid ⊢ ( invg ‘ 𝐺 ) = ( invg ‘ 𝐺 )
5 1 4 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ∈ 𝐵 )
6 5 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ∈ 𝐵 )
7 1 2 grpcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ∈ 𝐵 ) → ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) ∈ 𝐵 )
8 6 7 syld3an3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) ∈ 𝐵 )
9 1 2 4 3 grpsubval ⊢ ( ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) ∈ 𝐵 ∧ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ∈ 𝐵 ) → ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) − ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) + ( ( invg ‘ 𝐺 ) ‘ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) ) )
10 8 6 9 syl2anc ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) − ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) + ( ( invg ‘ 𝐺 ) ‘ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) ) )
11 1 2 3 grppncan ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ∈ 𝐵 ) → ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) − ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = 𝑋 )
12 6 11 syld3an3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) − ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = 𝑋 )
13 1 2 4 3 grpsubval ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 − 𝑌 ) = ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) )
14 13 3adant1 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 − 𝑌 ) = ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) )
15 14 eqcomd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = ( 𝑋 − 𝑌 ) )
16 1 4 grpinvinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = 𝑌 )
17 16 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) = 𝑌 )
18 15 17 oveq12d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) + ( ( invg ‘ 𝐺 ) ‘ ( ( invg ‘ 𝐺 ) ‘ 𝑌 ) ) ) = ( ( 𝑋 − 𝑌 ) + 𝑌 ) )
19 10 12 18 3eqtr3rd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 − 𝑌 ) + 𝑌 ) = 𝑋 )