Term Rewriting System R:
[x, y]
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MIN(s(x), s(y)) -> MIN(x, y)
MAX(s(x), s(y)) -> MAX(x, y)
-'(s(x), s(y)) -> -'(x, y)
GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
GCD(s(x), s(y)) -> -'(s(max(x, y)), s(min(x, y)))
GCD(s(x), s(y)) -> MAX(x, y)
GCD(s(x), s(y)) -> MIN(x, y)

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Nar


Dependency Pair:

MIN(s(x), s(y)) -> MIN(x, y)


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MIN(s(x), s(y)) -> MIN(x, y)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MIN(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Nar


Dependency Pair:


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Nar


Dependency Pair:

MAX(s(x), s(y)) -> MAX(x, y)


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MAX(s(x), s(y)) -> MAX(x, y)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MAX(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 6
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Nar


Dependency Pair:


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →DP Problem 4
Nar


Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(x, y)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 7
Dependency Graph
       →DP Problem 4
Nar


Dependency Pair:


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Narrowing Transformation


Dependency Pair:

GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
10 new Dependency Pairs are created:

GCD(s(x''), s(y'')) -> GCD(-(max(x'', y''), min(x'', y'')), s(min(x'', y'')))
GCD(s(x''), s(0)) -> GCD(-(s(x''), s(min(x'', 0))), s(min(x'', 0)))
GCD(s(0), s(y'')) -> GCD(-(s(y''), s(min(0, y''))), s(min(0, y'')))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(s(max(x'', y''))), s(min(s(x''), s(y'')))), s(min(s(x''), s(y''))))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(0)), s(min(x'', 0)))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(0)), s(min(0, y'')))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(s(min(x'', y'')))), s(min(s(x''), s(y''))))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(min(x'', 0))), s(0))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(min(0, y''))), s(0))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(min(s(x''), s(y'')))), s(s(min(x'', y''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Nar
           →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(min(s(x''), s(y'')))), s(s(min(x'', y''))))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(min(0, y''))), s(0))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(min(x'', 0))), s(0))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(s(min(x'', y'')))), s(min(s(x''), s(y''))))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(0)), s(min(0, y'')))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(0)), s(min(x'', 0)))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(s(max(x'', y''))), s(min(s(x''), s(y'')))), s(min(s(x''), s(y''))))
GCD(s(0), s(y'')) -> GCD(-(s(y''), s(min(0, y''))), s(min(0, y'')))
GCD(s(x''), s(0)) -> GCD(-(s(x''), s(min(x'', 0))), s(min(x'', 0)))
GCD(s(x''), s(y'')) -> GCD(-(max(x'', y''), min(x'', y'')), s(min(x'', y'')))


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:06 minutes