I'm struggling with this:Define P m : N → N by P m ( n ) = nm and m n as ( P m ) n ( 1 )Now prove k m + n = k m k n , and k m n = ( k m ) n and ( k n ) ( m n ) = ( k m ) n for all k , m , n ∈ NMuch appreciated! - I can get the first one using induction but the others are allusive...

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I&#039;m struggling with this:Define       P    m    :  N  →  N by       P    m    (  n  ) = nm and       m    n   as   (      P    m        )    n    (  1  )Now prove       k          m      +      n        =      k    m        k    n  , and       k          m      n        =  (      k    m        )    n   and   (      k    n    )  (      m    n    )  =  (  k  m      )    n   for all   k  ,  m  ,  n  ∈  NMuch appreciated! - I can get the first one using induction but the others are allusive...

Dayana Zuniga · Accepted Answer

All three identities are proved via Mathematical Induction.We first need to prove recursive identities.Using the above definition for exponents of the natural numbers it is shown that       m    0    =  1 and       m          S      (      n      )        =  m  (      m    n    ), where S is the successor function from Peano&#039;s postulates.      m    0    =  (      P    m        )    0    (  1  )  =  1  N  (  1  )  =  1, so       m    0    =  1      m          S      (      n      )        =  (      P    m        )          S      (      n      )        (  1  )  =      P    m    (      P    m        )    n    (  1  )  =      P    m    (      m    n    )  =  m  (      m    n    )Let   U be the set of natural numbers, for which       k          m      +      n        =      k    m        k    n   is true. We use induction on   n  ∈  N for all natural numbers   m.  0  ∈  U  , as       k          m      +      0        =      k    m    =      k    m    1  =      k    m        k    0  Assume   n  ∈  U, then       k          m      +      S      (      n      )        =      k          S      (      m      +      n      )        =  k  (      k          m      +      n        )  =  k  (      k    m        k    n    ) =   k  (      k    n        k    m    ) =   (  k      k    n    )      k    m   =       k          S      (      n      )            k    m  So   S  (  n  )  ∈  U and   U  =  N by PMISecond identity: Prove       k          m      n        =  (      k    m        )    n  The proofs of the other two identities are similar. Let me know if you would like them and I&#039;ll post them...

I'm struggling with this: Define P m </msub> : N →

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