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Metamath Proof Explorer

Theorem npncan

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005)

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

Proof

Step Hyp Ref Expression
1 subcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
2 1 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
3 addsubass ⊢ ( ( ( 𝐴 − 𝐵 ) ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) − 𝐶 ) = ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐶 ) ) )
4 2 3 syld3an1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) − 𝐶 ) = ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐶 ) ) )
5 npcan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + 𝐵 ) = 𝐴 )
6 5 oveq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) − 𝐶 ) = ( 𝐴 − 𝐶 ) )
7 6 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) − 𝐶 ) = ( 𝐴 − 𝐶 ) )
8 4 7 eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐶 ) ) = ( 𝐴 − 𝐶 ) )