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Metamath Proof Explorer

Theorem nppcan2d

Description: Cancellation law for subtraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses negidd.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
pncand.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
subaddd.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
Assertion nppcan2d ( 𝜑 → ( ( 𝐴 − ( 𝐵 + 𝐶 ) ) + 𝐶 ) = ( 𝐴 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 negidd.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 pncand.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 subaddd.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 nppcan2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 + 𝐶 ) ) + 𝐶 ) = ( 𝐴 − 𝐵 ) )
5 1 2 3 4 syl3anc ⊢ ( 𝜑 → ( ( 𝐴 − ( 𝐵 + 𝐶 ) ) + 𝐶 ) = ( 𝐴 − 𝐵 ) )