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# Assign1to3 - Assign#1 Show that prfs are closed under...

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Assign#1 Show that prfs are closed under Fibonacci induction. Fibonacci induction means that at each induction step, y+1, after calculating the two base values is computed using the y-th and (y-1)st values. The formal hypothesis is: Assume g1, g2 and h are already known to be prf, then so is f, where f(x,0) = g1(x); f(x,1) = g2(x) f(x,y+1) = h(f(x,y),f(x,y-1)) Hint: The pairing function is useful here. Let K be the following primitive recursive function, defined by induction on the primitive recursive functions, g, h, and the pairing function. K(x,0) = < g2(x), g1(x) > K(x, y+1) = J(x, y, K(y,x)) J(x,y,z) = < h(<z> 1 , <z> 2 ), <z> 1 > // this is < f(y+2,x), f(y+1,x)>, even though f is not yet shown to be prf!! This shows K is prf. f is then defined from K as follows: f(y,x) = <K(y,x)> 2 // extract second value from pair encoded in K(y,x) This shows it is also a prf, as was desired. Assign # 2 1. For the sets in a) and b), write a set description that involves the use of a minimum sequence of alternating quantifiers in front of a totally computable predicate (typically formed from STP
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