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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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), 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)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
LOG(s(s(x))) -> QUOT(x, s(s(0)))

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
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(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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


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

resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)
le(x1, x2) -> le(x1, x2)
minus(x1, x2) -> x1
ifminus(x1, x2, x3) -> x2
quot(x1, x2) -> x1
log(x1) -> log(x1)


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


Dependency Pair:


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





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
       →DP Problem 4
AFS


Dependency Pairs:

IFMINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x), y) -> IFMINUS(le(s(x), y), s(x), y)


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{log, s} > false > 0
{MINUS, IFMINUS} > 0
ifminus > 0
quot > 0
minus > 0
le > true > 0

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
IFMINUS(x1, x2, x3) -> IFMINUS(x2, x3)
s(x1) -> s(x1)
le(x1, x2) -> le(x1, x2)
minus(x1, x2) -> x1
ifminus(x1, x2, x3) -> x2
quot(x1, x2) -> x1
log(x1) -> log(x1)


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


Dependency Pair:

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


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





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
       →DP Problem 4
AFS


Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


The following rules can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
log > s > 0 > false
ifminus > false
quot > false
minus > false
QUOT > false
le > true > false

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1
ifminus(x1, x2, x3) -> x2
le(x1, x2) -> le(x1, x2)
quot(x1, x2) -> x1
log(x1) -> log(x1)


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


Dependency Pair:


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Argument Filtering and Ordering


Dependency Pair:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





The following dependency pair can be strictly oriented:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))


The following rules can be oriented:

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


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

resulting in one new DP problem.
Used Argument Filtering System:
LOG(x1) -> LOG(x1)
s(x1) -> s(x1)
quot(x1, x2) -> x1
minus(x1, x2) -> x1
ifminus(x1, x2, x3) -> x2
le(x1, x2) -> le(x1, x2)
log(x1) -> log(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
           →DP Problem 8
Dependency Graph


Dependency Pair:


Rules:


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)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:03 minutes