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


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, x, y) -> mod(minus(x, y), y)
ifmod(false, s(x), s(y)) -> s(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: 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
Nar


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


Strategy:

innermost




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
Nar


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, x, y) -> mod(minus(x, y), y)
ifmod(false, s(x), s(y)) -> s(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: 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
Nar


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

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


Strategy:

innermost




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

MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Narrowing Transformation


Dependency Pairs:

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


Strategy:

innermost




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

IFMOD(true, x, y) -> MOD(minus(x, y), y)
two new Dependency Pairs are created:

IFMOD(true, x'', 0) -> MOD(x'', 0)
IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 7
Instantiation Transformation


Dependency Pairs:

MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))
IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))


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


Strategy:

innermost




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

IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
two new Dependency Pairs are created:

IFMOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 8
Rewriting Transformation


Dependency Pairs:

IFMOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))


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


Strategy:

innermost




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

IFMOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
one new Dependency Pair is created:

IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 10
Argument Filtering and Ordering


Dependency Pairs:

IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{MOD, s} > 0 > true
{MOD, s} > IFMOD > true

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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 12
Dependency Graph


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 9
Rewriting Transformation


Dependency Pairs:

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


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


Strategy:

innermost




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

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 11
Argument Filtering and Ordering


Dependency Pairs:

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


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rules for innermost 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
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 13
Dependency Graph


Dependency Pair:

MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:04 minutes