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Metamath Proof Explorer

Theorem grpsubrcan

Description: Right cancellation law for group subtraction. (Contributed by NM, 31-Mar-2014)

Ref Expression
Hypotheses grpsubcl.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpsubcl.m ⊢ − = ( -g ‘ 𝐺 )
Assertion grpsubrcan ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) = ( 𝑌 − 𝑍 ) ↔ 𝑋 = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 grpsubcl.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpsubcl.m ⊢ − = ( -g ‘ 𝐺 )
3 eqid ⊢ ( +g ‘ 𝐺 ) = ( +g ‘ 𝐺 )
4 eqid ⊢ ( invg ‘ 𝐺 ) = ( invg ‘ 𝐺 )
5 1 3 4 2 grpsubval ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑋 − 𝑍 ) = ( 𝑋 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) )
6 5 3adant2 ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑋 − 𝑍 ) = ( 𝑋 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) )
7 1 3 4 2 grpsubval ⊢ ( ( 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑌 − 𝑍 ) = ( 𝑌 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) )
8 7 3adant1 ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑌 − 𝑍 ) = ( 𝑌 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) )
9 6 8 eqeq12d ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( ( 𝑋 − 𝑍 ) = ( 𝑌 − 𝑍 ) ↔ ( 𝑋 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) = ( 𝑌 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) ) )
10 9 adantl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) = ( 𝑌 − 𝑍 ) ↔ ( 𝑋 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) = ( 𝑌 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) ) )
11 simpl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝐺 ∈ Grp )
12 simpr1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
13 simpr2 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑌 ∈ 𝐵 )
14 1 4 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑍 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ∈ 𝐵 )
15 14 3ad2antr3 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ∈ 𝐵 )
16 1 3 grprcan ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ∈ 𝐵 ) ) → ( ( 𝑋 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) = ( 𝑌 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) ↔ 𝑋 = 𝑌 ) )
17 11 12 13 15 16 syl13anc ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) = ( 𝑌 ( +g ‘ 𝐺 ) ( ( invg ‘ 𝐺 ) ‘ 𝑍 ) ) ↔ 𝑋 = 𝑌 ) )
18 10 17 bitrd ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) = ( 𝑌 − 𝑍 ) ↔ 𝑋 = 𝑌 ) )