Metamath Proof Explorer


Theorem relexpreld

Description: The exponentiation of a relation is a relation. (Contributed by Drahflow, 12-Nov-2015) (Revised by RP, 30-May-2020) (Revised by AV, 12-Jul-2024)

Ref Expression
Hypotheses relexpreld.1
|- ( ph -> Rel R )
relexpreld.2
|- ( ph -> N e. NN0 )
Assertion relexpreld
|- ( ph -> Rel ( R ^r N ) )

Proof

Step Hyp Ref Expression
1 relexpreld.1
 |-  ( ph -> Rel R )
2 relexpreld.2
 |-  ( ph -> N e. NN0 )
3 2 adantr
 |-  ( ( ph /\ R e. _V ) -> N e. NN0 )
4 simpr
 |-  ( ( ph /\ R e. _V ) -> R e. _V )
5 1 adantr
 |-  ( ( ph /\ R e. _V ) -> Rel R )
6 relexprel
 |-  ( ( N e. NN0 /\ R e. _V /\ Rel R ) -> Rel ( R ^r N ) )
7 3 4 5 6 syl3anc
 |-  ( ( ph /\ R e. _V ) -> Rel ( R ^r N ) )
8 7 ex
 |-  ( ph -> ( R e. _V -> Rel ( R ^r N ) ) )
9 rel0
 |-  Rel (/)
10 reldmrelexp
 |-  Rel dom ^r
11 10 ovprc1
 |-  ( -. R e. _V -> ( R ^r N ) = (/) )
12 11 releqd
 |-  ( -. R e. _V -> ( Rel ( R ^r N ) <-> Rel (/) ) )
13 9 12 mpbiri
 |-  ( -. R e. _V -> Rel ( R ^r N ) )
14 8 13 pm2.61d1
 |-  ( ph -> Rel ( R ^r N ) )