Term Rewriting System R:
[X, Y, Z]
plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(plus(X, Y), Z) -> PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) -> PLUS(Y, Z)
TIMES(X, s(Y)) -> PLUS(X, times(Y, X))
TIMES(X, s(Y)) -> TIMES(Y, X)

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
FwdInst


Dependency Pair:

PLUS(plus(X, Y), Z) -> PLUS(Y, Z)


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

PLUS(plus(X, Y), Z) -> PLUS(Y, Z)
one new Dependency Pair is created:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')

The transformation is resulting in one new DP problem:



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


Dependency Pair:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




The following dependency pair can be strictly oriented:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')


The following usable rule for innermost can be oriented:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


Dependency Pair:


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Forward Instantiation Transformation


Dependency Pair:

TIMES(X, s(Y)) -> TIMES(Y, X)


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

TIMES(X, s(Y)) -> TIMES(Y, X)
one new Dependency Pair is created:

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 5
Forward Instantiation Transformation


Dependency Pair:

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))
one new Dependency Pair is created:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 6
Forward Instantiation Transformation


Dependency Pair:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))
one new Dependency Pair is created:

TIMES(s(s(Y'''''''')), s(s(Y'''''0))) -> TIMES(s(Y'''''0), s(s(Y'''''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 7
Forward Instantiation Transformation


Dependency Pair:

TIMES(s(s(Y'''''''')), s(s(Y'''''0))) -> TIMES(s(Y'''''0), s(s(Y'''''''')))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

TIMES(s(s(Y'''''''')), s(s(Y'''''0))) -> TIMES(s(Y'''''0), s(s(Y'''''''')))
one new Dependency Pair is created:

TIMES(s(s(Y'''''''''')), s(s(s(Y''''''''''')))) -> TIMES(s(s(Y''''''''''')), s(s(Y'''''''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 8
Forward Instantiation Transformation


Dependency Pair:

TIMES(s(s(Y'''''''''')), s(s(s(Y''''''''''')))) -> TIMES(s(s(Y''''''''''')), s(s(Y'''''''''')))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

TIMES(s(s(Y'''''''''')), s(s(s(Y''''''''''')))) -> TIMES(s(s(Y''''''''''')), s(s(Y'''''''''')))
one new Dependency Pair is created:

TIMES(s(s(s(Y''''''''''''''))), s(s(s(Y'''''''''''0)))) -> TIMES(s(s(Y'''''''''''0)), s(s(s(Y''''''''''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 9
Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

TIMES(s(s(s(Y''''''''''''''))), s(s(s(Y'''''''''''0)))) -> TIMES(s(s(Y'''''''''''0)), s(s(s(Y''''''''''''''))))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost



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