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Metamath Proof Explorer

Theorem npncan3

Description: Cancellation law for subtraction. (Contributed by Scott Fenton, 23-Jun-2013) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion npncan3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐴 ) ) = ( 𝐶 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 simp1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
2 subcl ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐶 − 𝐴 ) ∈ ℂ )
3 2 ancoms ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 − 𝐴 ) ∈ ℂ )
4 3 3adant2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 − 𝐴 ) ∈ ℂ )
5 simp2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
6 addsub ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐶 − 𝐴 ) ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + ( 𝐶 − 𝐴 ) ) − 𝐵 ) = ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐴 ) ) )
7 1 4 5 6 syl3anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + ( 𝐶 − 𝐴 ) ) − 𝐵 ) = ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐴 ) ) )
8 pncan3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + ( 𝐶 − 𝐴 ) ) = 𝐶 )
9 8 3adant2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + ( 𝐶 − 𝐴 ) ) = 𝐶 )
10 9 oveq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + ( 𝐶 − 𝐴 ) ) − 𝐵 ) = ( 𝐶 − 𝐵 ) )
11 7 10 eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐴 ) ) = ( 𝐶 − 𝐵 ) )