Metamath Proof Explorer


Theorem subcan2i

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

Ref Expression
Hypotheses negidi.1 𝐴 ∈ ℂ
pncan3i.2 𝐵 ∈ ℂ
subadd.3 𝐶 ∈ ℂ
Assertion subcan2i ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 )

Proof

Step Hyp Ref Expression
1 negidi.1 𝐴 ∈ ℂ
2 pncan3i.2 𝐵 ∈ ℂ
3 subadd.3 𝐶 ∈ ℂ
4 subcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 ) )
5 1 2 3 4 mp3an ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 )