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Metamath Proof Explorer

Theorem addsub

Description: Law for addition and subtraction. (Contributed by NM, 19-Aug-2001) (Proof shortened by Andrew Salmon, 22-Oct-2011)

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

Proof

Step Hyp Ref Expression
1 addcom ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
2 1 oveq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 𝐶 ) = ( ( 𝐵 + 𝐴 ) − 𝐶 ) )
3 2 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 𝐶 ) = ( ( 𝐵 + 𝐴 ) − 𝐶 ) )
4 addsubass ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − 𝐶 ) = ( 𝐵 + ( 𝐴 − 𝐶 ) ) )
5 4 3com12 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − 𝐶 ) = ( 𝐵 + ( 𝐴 − 𝐶 ) ) )
6 subcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐶 ) ∈ ℂ )
7 addcom ⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐴 − 𝐶 ) ∈ ℂ ) → ( 𝐵 + ( 𝐴 − 𝐶 ) ) = ( ( 𝐴 − 𝐶 ) + 𝐵 ) )
8 6 7 sylan2 ⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ) → ( 𝐵 + ( 𝐴 − 𝐶 ) ) = ( ( 𝐴 − 𝐶 ) + 𝐵 ) )
9 8 3impb ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + ( 𝐴 − 𝐶 ) ) = ( ( 𝐴 − 𝐶 ) + 𝐵 ) )
10 9 3com12 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + ( 𝐴 − 𝐶 ) ) = ( ( 𝐴 − 𝐶 ) + 𝐵 ) )
11 3 5 10 3eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 𝐶 ) = ( ( 𝐴 − 𝐶 ) + 𝐵 ) )