Computational Complexity.
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Show that prfs are closed under three-way mutual induction. Three-way mutual induction means that each induction step after calculating the base is computed using the previous value of the other function. The formal hypothesis is:
Assume g1, g2, g3, h1, h2, and h3are already known to be prf, then so are f1, f2, and f3, where
f1(x,0) = g1(x); f1(x,y+1) = h1(f2(x,y),f3(x,y));
f2(x,0) = g2(x); f2(x,y+1) = h2(f3(x,y),f1(x,y))
f3(x,0) = g3(x); f3(x,y+1) = h3(f1(x,y),f2(x,y))
Let S be an arbitrary non-empty, re set. Furthermore, let S be the range of some partial recursive function fs. Show that S is the range of some primitive recursive function, call it hs.
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