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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)

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)
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) -> LE(y, x)
IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
IFMOD(true, s(x), s(y)) -> MINUS(x, y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
AFS


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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 4
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
AFS


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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Argument Filtering and Ordering
       →DP Problem 3
AFS


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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 3
AFS


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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Argument Filtering and Ordering


Dependency Pairs:

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IFMOD(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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)





The following dependency pair can be strictly oriented:

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))


The following usable rules using the Ce-refinement can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{MOD, IFMOD}
{true, 0}
s > false

resulting in one new DP problem.
Used Argument Filtering System:
MOD(x1, x2) -> MOD(x1, x2)
IFMOD(x1, x2, x3) -> IFMOD(x2, x3)
s(x1) -> s(x1)
minus(x1, x2) -> x1
le(x1, x2) -> le(x1, x2)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
           →DP Problem 6
Dependency Graph


Dependency Pair:

MOD(s(x), s(y)) -> IFMOD(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(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes