# Metamath Proof Explorer

## Theorem subcan2

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

Ref Expression
Assertion subcan2
`|- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> A = B ) )`

### Proof

Step Hyp Ref Expression
1 simp1
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> A e. CC )`
2 simp3
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> C e. CC )`
3 subcl
` |-  ( ( B e. CC /\ C e. CC ) -> ( B - C ) e. CC )`
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( B - C ) e. CC )`
` |-  ( ( A e. CC /\ C e. CC /\ ( B - C ) e. CC ) -> ( ( A - C ) = ( B - C ) <-> ( ( B - C ) + C ) = A ) )`
6 1 2 4 5 syl3anc
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> ( ( B - C ) + C ) = A ) )`
7 npcan
` |-  ( ( B e. CC /\ C e. CC ) -> ( ( B - C ) + C ) = B )`
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( B - C ) + C ) = B )`
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( B - C ) + C ) = A <-> B = A ) )`
` |-  ( B = A <-> A = B )`
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( B - C ) + C ) = A <-> A = B ) )`
` |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> A = B ) )`