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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 ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 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 ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) )
5 1 2 3 4 mp3an ⊢ ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 )