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Metamath Proof Explorer

Theorem ppncan

Description: Cancellation law for mixed addition and subtraction. (Contributed by NM, 30-Jun-2005)

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

Proof

Step Hyp Ref Expression
1 addcom ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
2 1 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
3 2 oveq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐵 − 𝐶 ) ) = ( ( 𝐵 + 𝐴 ) − ( 𝐵 − 𝐶 ) ) )
4 addcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + 𝐵 ) ∈ ℂ )
5 4 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐵 ) ∈ ℂ )
6 subsub2 ⊢ ( ( ( 𝐴 + 𝐵 ) ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐵 − 𝐶 ) ) = ( ( 𝐴 + 𝐵 ) + ( 𝐶 − 𝐵 ) ) )
7 5 6 syld3an1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐵 − 𝐶 ) ) = ( ( 𝐴 + 𝐵 ) + ( 𝐶 − 𝐵 ) ) )
8 pnncan ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − ( 𝐵 − 𝐶 ) ) = ( 𝐴 + 𝐶 ) )
9 8 3com12 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − ( 𝐵 − 𝐶 ) ) = ( 𝐴 + 𝐶 ) )
10 3 7 9 3eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶 − 𝐵 ) ) = ( 𝐴 + 𝐶 ) )