Metamath Proof Explorer


Theorem 1arithlem1

Description: Lemma for 1arith . (Contributed by Mario Carneiro, 30-May-2014)

Ref Expression
Hypothesis 1arith.1
|- M = ( n e. NN |-> ( p e. Prime |-> ( p pCnt n ) ) )
Assertion 1arithlem1
|- ( N e. NN -> ( M ` N ) = ( p e. Prime |-> ( p pCnt N ) ) )

Proof

Step Hyp Ref Expression
1 1arith.1
 |-  M = ( n e. NN |-> ( p e. Prime |-> ( p pCnt n ) ) )
2 oveq2
 |-  ( n = N -> ( p pCnt n ) = ( p pCnt N ) )
3 2 mpteq2dv
 |-  ( n = N -> ( p e. Prime |-> ( p pCnt n ) ) = ( p e. Prime |-> ( p pCnt N ) ) )
4 prmex
 |-  Prime e. _V
5 4 mptex
 |-  ( p e. Prime |-> ( p pCnt N ) ) e. _V
6 3 1 5 fvmpt
 |-  ( N e. NN -> ( M ` N ) = ( p e. Prime |-> ( p pCnt N ) ) )