Metamath Proof Explorer


Theorem zsubcl

Description: Closure of subtraction of integers. (Contributed by NM, 11-May-2004)

Ref Expression
Assertion zsubcl
|- ( ( M e. ZZ /\ N e. ZZ ) -> ( M - N ) e. ZZ )

Proof

Step Hyp Ref Expression
1 zcn
 |-  ( M e. ZZ -> M e. CC )
2 zcn
 |-  ( N e. ZZ -> N e. CC )
3 negsub
 |-  ( ( M e. CC /\ N e. CC ) -> ( M + -u N ) = ( M - N ) )
4 1 2 3 syl2an
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( M + -u N ) = ( M - N ) )
5 znegcl
 |-  ( N e. ZZ -> -u N e. ZZ )
6 zaddcl
 |-  ( ( M e. ZZ /\ -u N e. ZZ ) -> ( M + -u N ) e. ZZ )
7 5 6 sylan2
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( M + -u N ) e. ZZ )
8 4 7 eqeltrrd
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( M - N ) e. ZZ )