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Metamath Proof Explorer

Theorem subsub4

Description: Law for double subtraction. (Contributed by NM, 19-Aug-2005) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion subsub4 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − 𝐶 ) = ( 𝐴 − ( 𝐵 + 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 nppcan2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 + 𝐶 ) ) + 𝐶 ) = ( 𝐴 − 𝐵 ) )
2 simp1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
3 simp2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
4 subcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
5 2 3 4 syl2anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
6 simp3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
7 3 6 addcld ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) ∈ ℂ )
8 subcl ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 + 𝐶 ) ∈ ℂ ) → ( 𝐴 − ( 𝐵 + 𝐶 ) ) ∈ ℂ )
9 2 7 8 syl2anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐵 + 𝐶 ) ) ∈ ℂ )
10 subadd2 ⊢ ( ( ( 𝐴 − 𝐵 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ ( 𝐴 − ( 𝐵 + 𝐶 ) ) ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) − 𝐶 ) = ( 𝐴 − ( 𝐵 + 𝐶 ) ) ↔ ( ( 𝐴 − ( 𝐵 + 𝐶 ) ) + 𝐶 ) = ( 𝐴 − 𝐵 ) ) )
11 5 6 9 10 syl3anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) − 𝐶 ) = ( 𝐴 − ( 𝐵 + 𝐶 ) ) ↔ ( ( 𝐴 − ( 𝐵 + 𝐶 ) ) + 𝐶 ) = ( 𝐴 − 𝐵 ) ) )
12 1 11 mpbird ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − 𝐶 ) = ( 𝐴 − ( 𝐵 + 𝐶 ) ) )