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Metamath Proof Explorer

Theorem nnncan2

Description: Cancellation law for subtraction. (Contributed by NM, 1-Oct-2005)

Ref Expression
Assertion nnncan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐶 ) ) = ( 𝐴 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 subcl ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
2 1 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
3 sub32 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 − 𝐶 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐶 ) ) )
4 2 3 syld3an2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐶 ) ) )
5 nnncan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( 𝐴 − 𝐵 ) )
6 4 5 eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐶 ) ) = ( 𝐴 − 𝐵 ) )