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Metamath Proof Explorer

Theorem ablsubsub

Description: Law for double subtraction. (Contributed by NM, 7-Apr-2015)

Ref Expression
Hypotheses ablsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
ablsubadd.p ⊢ + = ( +g ‘ 𝐺 )
ablsubadd.m ⊢ − = ( -g ‘ 𝐺 )
ablsubsub.g ⊢ ( 𝜑 → 𝐺 ∈ Abel )
ablsubsub.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
ablsubsub.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
ablsubsub.z ⊢ ( 𝜑 → 𝑍 ∈ 𝐵 )
Assertion ablsubsub ( 𝜑 → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( ( 𝑋 − 𝑌 ) + 𝑍 ) )

Proof

Step Hyp Ref Expression
1 ablsubadd.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 ablsubadd.p ⊢ + = ( +g ‘ 𝐺 )
3 ablsubadd.m ⊢ − = ( -g ‘ 𝐺 )
4 ablsubsub.g ⊢ ( 𝜑 → 𝐺 ∈ Abel )
5 ablsubsub.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
6 ablsubsub.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
7 ablsubsub.z ⊢ ( 𝜑 → 𝑍 ∈ 𝐵 )
8 ablgrp ⊢ ( 𝐺 ∈ Abel → 𝐺 ∈ Grp )
9 4 8 syl ⊢ ( 𝜑 → 𝐺 ∈ Grp )
10 1 2 3 grpsubsub ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )
11 9 5 6 7 10 syl13anc ⊢ ( 𝜑 → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )
12 1 2 3 grpaddsubass ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 + 𝑍 ) − 𝑌 ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )
13 9 5 7 6 12 syl13anc ⊢ ( 𝜑 → ( ( 𝑋 + 𝑍 ) − 𝑌 ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )
14 1 2 3 abladdsub ⊢ ( ( 𝐺 ∈ Abel ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 + 𝑍 ) − 𝑌 ) = ( ( 𝑋 − 𝑌 ) + 𝑍 ) )
15 4 5 7 6 14 syl13anc ⊢ ( 𝜑 → ( ( 𝑋 + 𝑍 ) − 𝑌 ) = ( ( 𝑋 − 𝑌 ) + 𝑍 ) )
16 11 13 15 3eqtr2d ⊢ ( 𝜑 → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( ( 𝑋 − 𝑌 ) + 𝑍 ) )