Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, x, y) -> gcd(minus(x, y), y)
ifgcd(false, x, y) -> gcd(minus(y, x), x)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), s(y)) -> MINUS(x, y)
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IFGCD(true, x, y) -> GCD(minus(x, y), y)
IFGCD(true, x, y) -> MINUS(x, y)
IFGCD(false, x, y) -> GCD(minus(y, x), x)
IFGCD(false, x, y) -> MINUS(y, x)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Remaining


Dependency Pair:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, x, y) -> gcd(minus(x, y), y)
ifgcd(false, x, y) -> gcd(minus(y, x), x)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.



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


Dependency Pair:


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, x, y) -> gcd(minus(x, y), y)
ifgcd(false, x, y) -> gcd(minus(y, x), x)


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
Remaining


Dependency Pair:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, x, y) -> gcd(minus(x, y), y)
ifgcd(false, x, y) -> gcd(minus(y, x), x)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(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 5
Dependency Graph
       →DP Problem 3
Remaining


Dependency Pair:


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, x, y) -> gcd(minus(x, y), y)
ifgcd(false, x, y) -> gcd(minus(y, x), x)


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
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

IFGCD(false, x, y) -> GCD(minus(y, x), x)
IFGCD(true, x, y) -> GCD(minus(x, y), y)
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, x, y) -> gcd(minus(x, y), y)
ifgcd(false, x, y) -> gcd(minus(y, x), x)


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes