Term Rewriting System R:
[x, y, z]
times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

TIMES(x, plus(y, s(z))) -> PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) -> PLUS(y, times(s(z), 0))
TIMES(x, plus(y, s(z))) -> TIMES(s(z), 0)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, s(y)) -> PLUS(times(x, y), x)
TIMES(x, s(y)) -> TIMES(x, y)
PLUS(x, s(y)) -> PLUS(x, y)

Furthermore, R contains two SCCs.


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


Dependency Pair:

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


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


Strategy:

innermost




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

PLUS(x, s(y)) -> PLUS(x, y)
one new Dependency Pair is created:

PLUS(x'', s(s(y''))) -> PLUS(x'', s(y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
Forward Instantiation Transformation
       →DP Problem 2
FwdInst


Dependency Pair:

PLUS(x'', s(s(y''))) -> PLUS(x'', s(y''))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


Strategy:

innermost




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

PLUS(x'', s(s(y''))) -> PLUS(x'', s(y''))
one new Dependency Pair is created:

PLUS(x'''', s(s(s(y'''')))) -> PLUS(x'''', s(s(y'''')))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

PLUS(x'''', s(s(s(y'''')))) -> PLUS(x'''', s(s(y'''')))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

PLUS(x'''', s(s(s(y'''')))) -> PLUS(x'''', s(s(y'''')))


There are no usable rules for innermost w.r.t. to the AFS 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:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 3
FwdInst
             ...
               →DP Problem 5
Dependency Graph
       →DP Problem 2
FwdInst


Dependency Pair:


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


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(x, y)


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


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(x, y)
one new Dependency Pair is created:

TIMES(x'', s(s(y''))) -> TIMES(x'', s(y''))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

TIMES(x'', s(s(y''))) -> TIMES(x'', s(y''))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


Strategy:

innermost




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

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

TIMES(x'''', s(s(s(y'''')))) -> TIMES(x'''', s(s(y'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 6
FwdInst
             ...
               →DP Problem 7
Argument Filtering and Ordering


Dependency Pair:

TIMES(x'''', s(s(s(y'''')))) -> TIMES(x'''', s(s(y'''')))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

TIMES(x'''', s(s(s(y'''')))) -> TIMES(x'''', s(s(y'''')))


There are no usable rules for innermost w.r.t. to the AFS 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:
TIMES(x1, x2) -> TIMES(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 6
FwdInst
             ...
               →DP Problem 8
Dependency Graph


Dependency Pair:


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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