Metamath Proof Explorer


Theorem subcan2d

Description: Cancellation law for subtraction. (Contributed by Mario Carneiro, 22-Sep-2016)

Ref Expression
Hypotheses negidd.1 ( 𝜑𝐴 ∈ ℂ )
pncand.2 ( 𝜑𝐵 ∈ ℂ )
subaddd.3 ( 𝜑𝐶 ∈ ℂ )
subcan2d.4 ( 𝜑 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
Assertion subcan2d ( 𝜑𝐴 = 𝐵 )

Proof

Step Hyp Ref Expression
1 negidd.1 ( 𝜑𝐴 ∈ ℂ )
2 pncand.2 ( 𝜑𝐵 ∈ ℂ )
3 subaddd.3 ( 𝜑𝐶 ∈ ℂ )
4 subcan2d.4 ( 𝜑 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
5 subcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 ) )
6 1 2 3 5 syl3anc ( 𝜑 → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 ) )
7 4 6 mpbid ( 𝜑𝐴 = 𝐵 )