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Metamath Proof Explorer

Theorem grpsubsub

Description: Double group subtraction. (Contributed by NM, 24-Feb-2008) (Revised by Mario Carneiro, 2-Dec-2014)

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

Proof

Step Hyp Ref Expression
1 grpsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpsubadd.p ⊢ + = ( +g ‘ 𝐺 )
3 grpsubadd.m ⊢ − = ( -g ‘ 𝐺 )
4 simpr1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
5 1 3 grpsubcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑌 − 𝑍 ) ∈ 𝐵 )
6 5 3adant3r1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑌 − 𝑍 ) ∈ 𝐵 )
7 eqid ⊢ ( invg ‘ 𝐺 ) = ( invg ‘ 𝐺 )
8 1 2 7 3 grpsubval ⊢ ( ( 𝑋 ∈ 𝐵 ∧ ( 𝑌 − 𝑍 ) ∈ 𝐵 ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) ) )
9 4 6 8 syl2anc ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) ) )
10 1 3 7 grpinvsub ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) = ( 𝑍 − 𝑌 ) )
11 10 3adant3r1 ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) = ( 𝑍 − 𝑌 ) )
12 11 oveq2d ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )
13 9 12 eqtrd ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )