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Metamath Proof Explorer

Theorem subcan

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion subcan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) )

Proof

Step Hyp Ref Expression
1 simp2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
2 simp1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
3 1 2 addcomd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐵 ) )
4 3 eqeq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐶 ) ↔ ( 𝐴 + 𝐵 ) = ( 𝐴 + 𝐶 ) ) )
5 simp3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
6 addsubeq4 ⊢ ( ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) ∧ ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ) → ( ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐶 ) ↔ ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ) )
7 1 2 2 5 6 syl22anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐶 ) ↔ ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ) )
8 addcan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) = ( 𝐴 + 𝐶 ) ↔ 𝐵 = 𝐶 ) )
9 4 7 8 3bitr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) )