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Metamath Proof Explorer

Theorem ablnnncan1

Description: Cancellation law for group subtraction. ( nnncan1 analog.) (Contributed by NM, 7-Apr-2015)

Ref Expression
Hypotheses ablnncan.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
ablnncan.m ⊢ − = ( -g ‘ 𝐺 )
ablnncan.g ⊢ ( 𝜑 → 𝐺 ∈ Abel )
ablnncan.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
ablnncan.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
ablsub32.z ⊢ ( 𝜑 → 𝑍 ∈ 𝐵 )
Assertion ablnnncan1 ( 𝜑 → ( ( 𝑋 − 𝑌 ) − ( 𝑋 − 𝑍 ) ) = ( 𝑍 − 𝑌 ) )

Proof

Step Hyp Ref Expression
1 ablnncan.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 ablnncan.m ⊢ − = ( -g ‘ 𝐺 )
3 ablnncan.g ⊢ ( 𝜑 → 𝐺 ∈ Abel )
4 ablnncan.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 ablnncan.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
6 ablsub32.z ⊢ ( 𝜑 → 𝑍 ∈ 𝐵 )
7 ablgrp ⊢ ( 𝐺 ∈ Abel → 𝐺 ∈ Grp )
8 3 7 syl ⊢ ( 𝜑 → 𝐺 ∈ Grp )
9 1 2 grpsubcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑋 − 𝑍 ) ∈ 𝐵 )
10 8 4 6 9 syl3anc ⊢ ( 𝜑 → ( 𝑋 − 𝑍 ) ∈ 𝐵 )
11 1 2 3 4 5 10 ablsub32 ⊢ ( 𝜑 → ( ( 𝑋 − 𝑌 ) − ( 𝑋 − 𝑍 ) ) = ( ( 𝑋 − ( 𝑋 − 𝑍 ) ) − 𝑌 ) )
12 1 2 3 4 6 ablnncan ⊢ ( 𝜑 → ( 𝑋 − ( 𝑋 − 𝑍 ) ) = 𝑍 )
13 12 oveq1d ⊢ ( 𝜑 → ( ( 𝑋 − ( 𝑋 − 𝑍 ) ) − 𝑌 ) = ( 𝑍 − 𝑌 ) )
14 11 13 eqtrd ⊢ ( 𝜑 → ( ( 𝑋 − 𝑌 ) − ( 𝑋 − 𝑍 ) ) = ( 𝑍 − 𝑌 ) )