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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 )) . DeFne H iteratively as H (0 , x ) = a F 1 ( x ) , F 2 ( x ) A H ( y + 1 , x ) = a G 1 ( y, x, a H ( y, x ) A 2 ) , G 2 ( y, x, a H ( y, x ) A 1 ) A . 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 ) = a H 1 ( y, x ) , H 2 ( y, x ) A for all y 0. Base : H (0 , x ) = a F 1 ( x ) , F 2 ( x ) A = a H 1 (0 , x ) , H 2 (0 , x ) A . Hypothesis : H ( y, x ) = a H 1 ( y, x ) , H 2 ( y, x ) A for all y 0. Step : Let y 0. Then H ( y
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 07/14/2011 for the course COT 5310 taught by Professor Staff during the Spring '08 term at University of Central Florida.

Ask a homework question - tutors are online