Metamath Proof Explorer


Theorem subcani

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 subcani ( ( 𝐴𝐵 ) = ( 𝐴𝐶 ) ↔ 𝐵 = 𝐶 )

Proof

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