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Metamath Proof Explorer

Theorem ablsubadd

Description: Relationship between Abelian group subtraction and addition. (Contributed by NM, 31-Mar-2014)

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

Proof

Step Hyp Ref Expression
1 ablsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 ablsubadd.p ⊢ + = ( +g ‘ 𝐺 )
3 ablsubadd.m ⊢ − = ( -g ‘ 𝐺 )
4 ablgrp ⊢ ( 𝐺 ∈ Abel → 𝐺 ∈ Grp )
5 1 2 3 grpsubadd ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) = 𝑍 ↔ ( 𝑍 + 𝑌 ) = 𝑋 ) )
6 4 5 sylan ⊢ ( ( 𝐺 ∈ Abel ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) = 𝑍 ↔ ( 𝑍 + 𝑌 ) = 𝑋 ) )
7 1 2 ablcom ⊢ ( ( 𝐺 ∈ Abel ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑌 + 𝑍 ) = ( 𝑍 + 𝑌 ) )
8 7 3adant3r1 ⊢ ( ( 𝐺 ∈ Abel ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑌 + 𝑍 ) = ( 𝑍 + 𝑌 ) )
9 8 eqeq1d ⊢ ( ( 𝐺 ∈ Abel ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑌 + 𝑍 ) = 𝑋 ↔ ( 𝑍 + 𝑌 ) = 𝑋 ) )
10 6 9 bitr4d ⊢ ( ( 𝐺 ∈ Abel ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 − 𝑌 ) = 𝑍 ↔ ( 𝑌 + 𝑍 ) = 𝑋 ) )