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Metamath Proof Explorer

Theorem pnpcan2

Description: Cancellation law for mixed addition and subtraction. (Contributed by Scott Fenton, 9-Jun-2006)

Ref Expression
Assertion pnpcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( 𝐴 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 addcom ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) = ( 𝐶 + 𝐴 ) )
2 1 3adant2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) = ( 𝐶 + 𝐴 ) )
3 addcom ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
4 3 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
5 2 4 oveq12d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) )
6 pnpcan ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) = ( 𝐴 − 𝐵 ) )
7 6 3coml ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) = ( 𝐴 − 𝐵 ) )
8 5 7 eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( 𝐴 − 𝐵 ) )