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))))))

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), 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
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
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(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))))))


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: Homeomorphic Embedding Order with EMB
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 5
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


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))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Narrowing Transformation
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


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))))))


Strategy:

innermost




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

MINUS(s(x), y) -> IFMINUS(le(s(x), y), s(x), y)
two new Dependency Pairs are created:

MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IFMINUS(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 6
Instantiation Transformation
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

MINUS(s(x''), s(y'')) -> IFMINUS(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)
IFMINUS(false, s(x), y) -> MINUS(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))))))


Strategy:

innermost




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

IFMINUS(false, s(x), y) -> MINUS(x, y)
two new Dependency Pairs are created:

IFMINUS(false, s(x'), 0) -> MINUS(x', 0)
IFMINUS(false, s(x'), s(y'''')) -> MINUS(x', s(y''''))

The transformation is resulting in two new DP problems:



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




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 6
Inst
             ...
               →DP Problem 8
Argument Filtering and Ordering
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

IFMINUS(false, s(x'), 0) -> MINUS(x', 0)
MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 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))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFMINUS(false, s(x'), 0) -> MINUS(x', 0)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
IFMINUS(x1, x2, x3) -> x2
s(x1) -> s(x1)
MINUS(x1, x2) -> x1


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 6
Inst
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:

MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 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))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:


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




The following remains to be proven:

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