Metamath Proof Explorer


Theorem grpnpncan0

Description: Cancellation law for group subtraction ( npncan2 analog). (Contributed by AV, 24-Nov-2019)

Ref Expression
Hypotheses grpsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpsubadd.p ⊢ + = ( +g ‘ 𝐺 )
grpsubadd.m ⊢ − = ( -g ‘ 𝐺 )
grpnpncan0.0 ⊢ 0 = ( 0g ‘ 𝐺 )
Assertion grpnpncan0 ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) + ( 𝑌 − 𝑋 ) ) = 0 )

Proof

Step Hyp Ref Expression
1 grpsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpsubadd.p ⊢ + = ( +g ‘ 𝐺 )
3 grpsubadd.m ⊢ − = ( -g ‘ 𝐺 )
4 grpnpncan0.0 ⊢ 0 = ( 0g ‘ 𝐺 )
5 simpl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝐺 ∈ Grp )
6 simprl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
7 simprr ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑌 ∈ 𝐵 )
8 1 2 3 grpnpncan ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) + ( 𝑌 − 𝑋 ) ) = ( 𝑋 − 𝑋 ) )
9 5 6 7 6 8 syl13anc ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) + ( 𝑌 − 𝑋 ) ) = ( 𝑋 − 𝑋 ) )
10 1 4 3 grpsubid ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 − 𝑋 ) = 0 )
11 10 adantrr ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝑋 − 𝑋 ) = 0 )
12 9 11 eqtrd ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) + ( 𝑌 − 𝑋 ) ) = 0 )