Term Rewriting System R:

   [Y, X]
   le(0, Y) -> true
   le(s(X), 0) -> false
   le(s(X), s(Y)) -> le(X, Y)
   minus(0, Y) -> 0
   minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
   ifMinus(true, s(X), Y) -> 0
   ifMinus(false, s(X), Y) -> s(minus(X, Y))
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Termination of R to be shown.


This program has no overlaps, so it is sufficient to show innermost termination. 


R contains the following Dependency Pairs: 


   MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
   MINUS(s(X), Y) -> LE(s(X), Y)
   IFMINUS(false, s(X), Y) -> MINUS(X, Y)
   LE(s(X), s(Y)) -> LE(X, Y)
   QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
   QUOT(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.


SCC1:

LE(s(X), s(Y)) -> LE(X, Y)

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(LE(x_1, x_2)) = 1 + x_1 + x_2
POL(s(x_1)) = 1 + x_1

The following Dependency Pairs can be deleted:

   LE(s(X), s(Y)) -> LE(X, Y)



 This transformation is resulting in no new subcycles.


SCC2:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

By using a polynomial ordering, at least one Dependency Pair of this SCC can be strictly oriented.

No rules need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(le(x_1, x_2)) = 0
POL(true) = 0
POL(IFMINUS(x_1, x_2, x_3)) = x_2
POL(MINUS(x_1, x_2)) = x_1
POL(0) = 0
POL(false) = 0

 resulting in no subcycles.


SCC3:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

By using a polynomial ordering, at least one Dependency Pair of this SCC can be strictly oriented.

Additionally, the following rules can be oriented: 

   ifMinus(true, s(X), Y) -> 0
   ifMinus(false, s(X), Y) -> s(minus(X, Y))
   minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
   minus(0, Y) -> 0


Used ordering: Polynomial ordering with Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(le(x_1, x_2)) = 0
POL(ifMinus(x_1, x_2, x_3)) = x_2
POL(minus(x_1, x_2)) = x_1
POL(true) = 0
POL(QUOT(x_1, x_2)) = x_1 + x_2
POL(0) = 0
POL(false) = 0

 resulting in no subcycles.




Termination of R successfully shown.


Duration: 0.838 seconds.