Theorem subcan2

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005)

	Ref	Expression
Assertion	subcan2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		simp1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
2		simp3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
3		subcl	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
4	3	3adant1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
5		subadd2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ ( 𝐵 − 𝐶 ) ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ) )
6	1 2 4 5	syl3anc	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ) )
7		npcan	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐵 )
8	7	3adant1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐵 )
9	8	eqeq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ↔ 𝐵 = 𝐴 ) )
10		eqcom	⊢ ( 𝐵 = 𝐴 ↔ 𝐴 = 𝐵 )
11	9 10	syl6bb	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ↔ 𝐴 = 𝐵 ) )
12	6 11	bitrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ 𝐴 = 𝐵 ) )