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Metamath Proof Explorer

Theorem grpnnncan2

Description: Cancellation law for group subtraction. ( nnncan2 analog.) (Contributed by NM, 15-Feb-2008) (Revised by Mario Carneiro, 2-Dec-2014)

Ref Expression
Hypotheses grpnnncan2.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpnnncan2.m ⊢ − = ( -g ‘ 𝐺 )
Assertion grpnnncan2 ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) − ( 𝑌 − 𝑍 ) ) = ( 𝑋 − 𝑌 ) )

Proof

Step Hyp Ref Expression
1 grpnnncan2.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpnnncan2.m ⊢ − = ( -g ‘ 𝐺 )
3 simpl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝐺 ∈ Grp )
4 simpr1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
5 simpr3 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑍 ∈ 𝐵 )
6 1 2 grpsubcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑌 − 𝑍 ) ∈ 𝐵 )
7 6 3adant3r1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑌 − 𝑍 ) ∈ 𝐵 )
8 eqid ⊢ ( +g ‘ 𝐺 ) = ( +g ‘ 𝐺 )
9 1 8 2 grpsubsub4 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ∧ ( 𝑌 − 𝑍 ) ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) − ( 𝑌 − 𝑍 ) ) = ( 𝑋 − ( ( 𝑌 − 𝑍 ) ( +g ‘ 𝐺 ) 𝑍 ) ) )
10 3 4 5 7 9 syl13anc ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) − ( 𝑌 − 𝑍 ) ) = ( 𝑋 − ( ( 𝑌 − 𝑍 ) ( +g ‘ 𝐺 ) 𝑍 ) ) )
11 1 8 2 grpnpcan ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( ( 𝑌 − 𝑍 ) ( +g ‘ 𝐺 ) 𝑍 ) = 𝑌 )
12 11 3adant3r1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑌 − 𝑍 ) ( +g ‘ 𝐺 ) 𝑍 ) = 𝑌 )
13 12 oveq2d ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( ( 𝑌 − 𝑍 ) ( +g ‘ 𝐺 ) 𝑍 ) ) = ( 𝑋 − 𝑌 ) )
14 10 13 eqtrd ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑍 ) − ( 𝑌 − 𝑍 ) ) = ( 𝑋 − 𝑌 ) )