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Metamath Proof Explorer

Theorem grppncan

Description: Cancellation law for subtraction ( pncan analog). (Contributed by NM, 16-Apr-2014)

Ref Expression
Hypotheses grpsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpsubadd.p ⊢ + = ( +g ‘ 𝐺 )
grpsubadd.m ⊢ − = ( -g ‘ 𝐺 )
Assertion grppncan ( ( 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 simp1 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝐺 ∈ Grp )
5 simp2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 )
6 simp3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑌 ∈ 𝐵 )
7 1 2 3 grpaddsubass ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 + 𝑌 ) − 𝑌 ) = ( 𝑋 + ( 𝑌 − 𝑌 ) ) )
8 4 5 6 6 7 syl13anc ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑌 ) = ( 𝑋 + ( 𝑌 − 𝑌 ) ) )
9 eqid ⊢ ( 0g ‘ 𝐺 ) = ( 0g ‘ 𝐺 )
10 1 9 3 grpsubid ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 − 𝑌 ) = ( 0g ‘ 𝐺 ) )
11 10 oveq2d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 𝑌 − 𝑌 ) ) = ( 𝑋 + ( 0g ‘ 𝐺 ) ) )
12 11 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 𝑌 − 𝑌 ) ) = ( 𝑋 + ( 0g ‘ 𝐺 ) ) )
13 1 2 9 grprid ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 + ( 0g ‘ 𝐺 ) ) = 𝑋 )
14 13 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 0g ‘ 𝐺 ) ) = 𝑋 )
15 8 12 14 3eqtrd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑌 ) = 𝑋 )