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
Usable Rules (Innermost)
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules


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




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 5
Size-Change Principle
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LE(s(x), s(y)) -> LE(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules


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




As we are in the innermost case, we can delete all 8 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 6
Size-Change Principle
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




We number the DPs as follows:
  1. IFMINUS(false, s(x), y) -> MINUS(x, y)
  2. MINUS(s(x), y) -> IFMINUS(le(s(x), y), s(x), y)
and get the following Size-Change Graph(s):
{1} , {1}
2>1
3=2
{2} , {2}
1=2
2=3

which lead(s) to this/these maximal multigraph(s):
{1} , {2}
2>2
3=3
{2} , {1}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
Usable Rules (Innermost)
       →DP Problem 4
UsableRules


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))))))


Strategy:

innermost




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 7
Negative Polynomial Order
       →DP Problem 4
UsableRules


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following Dependency Pair can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

POL( le(x1, x2) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( ifminus(x1, ..., x3) ) = x2

POL( 0 ) = 0


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 7
Neg POLO
             ...
               →DP Problem 8
Dependency Graph
       →DP Problem 4
UsableRules


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
Usable Rules (Innermost)


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))))))


Strategy:

innermost




As we are in the innermost case, we can delete all 2 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
           →DP Problem 9
Negative Polynomial Order


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following Dependency Pair can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( LOG(x1) ) = x1

POL( s(x1) ) = x1 + 1

POL( quot(x1, x2) ) = x1

POL( le(x1, x2) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( 0 ) = 0

POL( minus(x1, x2) ) = x1

POL( ifminus(x1, ..., x3) ) = x2


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
           →DP Problem 9
Neg POLO
             ...
               →DP Problem 10
Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes