COT5310hw3key

COT5310hw3key - 1 x = a G 1 y x a H y x A 2 G 2 y x a H y x...

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COT 5310 Homework 3 Key 3. Show that prfs are closed under mutual recursion. That is, assuming F 1 , F 2 and G 1 and G 2 are pr, show that H 1 and H 2 are, where H 1 (0 ,x ) = F 1 ( x ) H 1 ( y + 1 ,x ) = G 1 ( y,x,H 2 ( y,x )) H 2 (0 ,x ) = F 2 ( x ) H 2 ( y + 1 ,x ) = G 2 ( y,x,H 1 ( y,x )) . Define H iteratively as H (0 ,x ) = ( F 1 ( x ) ,F 2 ( x ) ) H ( y + 1 ,x ) = ( G 1 ( y,x, ( H ( y,x ) ) 2 ) ,G 2 ( y,x, ( H ( y,x ) ) 1 ) ) . H is primitive recursive by construction since pairing, G 1 , G 2 , F 1 , and F 2 are all primitive recursive. The claim we need to prove (via induction) is that H ( y,x ) = ( H 1 ( y,x ) ,H 2 ( y,x ) ) for all y 0. Base : H (0 ,x ) = ( F 1 ( x ) ,F 2 ( x ) ) = ( H 1 (0 ,x ) ,H 2 (0 ,x ) ) . Hypothesis : H ( y,x ) = ( H 1 ( y,x ) ,H 2 ( y,x ) ) for all y 0. Step : Let y 0. Then H ( y + 1
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Unformatted text preview: + 1 , x ) = a G 1 ( y, x, a H ( y, x ) A 2 ) , G 2 ( y, x, a H ( y, x ) A 1 ) A . . . by deFnition H ( y + 1 , x ) = a G 1 ( y, x, H 2 ( y, x )) , G 2 ( y, x, H 1 ( y, x )) A . . . by hypothesis H ( y + 1 , x ) = a H 1 ( y + 1 , x ) , H 2 ( y + 1 , x ) A . . . deFnition of H 1 and H 2 . Now, H 1 ( y, x ) = a H ( y, x ) A 1 so H 1 is primitive recursive, and H 2 ( y, x ) = a H ( y, x ) A 2 so H 2 is primitive recursive. 1...
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