Term Rewriting System R:
[f, x, y, z]
app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) -> APP(plus, x)
APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
APP(app(times, app(s, x)), y) -> APP(plus, app(app(times, x), y))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(times, app(s, x)), y) -> APP(times, x)
SUM -> APP(app(fold, add), 0)
SUM -> APP(fold, add)
PROD -> APP(app(fold, mul), app(s, 0))
PROD -> APP(fold, mul)
PROD -> APP(s, 0)

Furthermore, R contains three SCCs.


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


Dependency Pair:

APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(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 4
A-Transformation
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)


Rule:

none


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


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


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. PLUS(s(x), y) -> PLUS(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


Dependency Pair:

APP(app(times, app(s, x)), y) -> APP(app(times, x), y)


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(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
A-Transformation
       →DP Problem 3
UsableRules


Dependency Pair:

APP(app(times, app(s, x)), y) -> APP(app(times, x), y)


Rule:

none


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 6
ATrans
             ...
               →DP Problem 7
Size-Change Principle
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. TIMES(s(x), y) -> TIMES(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
UsableRules
       →DP Problem 3
Usable Rules (Innermost)


Dependency Pairs:

APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(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 8
Narrowing Transformation


Dependency Pairs:

APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(plus, 0), y) -> y


Strategy:

innermost




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

APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
eight new Dependency Pairs are created:

APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, nil), z)) -> APP(x'', app(app(app(fold, app(app(fold, f''), x'')), x), z))
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(f'', y''), app(app(app(fold, f''), x''), z'')), app(app(app(fold, app(app(fold, f''), x'')), x), z))
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(s, app(app(plus, x''), y'')), app(app(app(fold, app(plus, app(s, x''))), x), z))
APP(app(app(fold, app(times, 0)), x), app(app(cons, y''), z)) -> APP(0, app(app(app(fold, app(times, 0)), x), z))
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(app(plus, app(app(times, x''), y'')), y''), app(app(app(fold, app(times, app(s, x''))), x), z))
APP(app(app(fold, app(plus, 0)), x), app(app(cons, y''), z)) -> APP(y'', app(app(app(fold, app(plus, 0)), x), z))
APP(app(app(fold, f''), x''), app(app(cons, y), nil)) -> APP(app(f'', y), x'')
APP(app(app(fold, f''), x''), app(app(cons, y), app(app(cons, y''), z''))) -> APP(app(f'', y), app(app(f'', y''), app(app(app(fold, f''), x''), z'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 8
Nar
             ...
               →DP Problem 9
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(app(app(fold, f''), x''), app(app(cons, y), app(app(cons, y''), z''))) -> APP(app(f'', y), app(app(f'', y''), app(app(app(fold, f''), x''), z'')))
APP(app(app(fold, f''), x''), app(app(cons, y), nil)) -> APP(app(f'', y), x'')
APP(app(app(fold, app(plus, 0)), x), app(app(cons, y''), z)) -> APP(y'', app(app(app(fold, app(plus, 0)), x), z))
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(app(plus, app(app(times, x''), y'')), y''), app(app(app(fold, app(times, app(s, x''))), x), z))
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(f'', y''), app(app(app(fold, f''), x''), z'')), app(app(app(fold, app(app(fold, f''), x'')), x), z))
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, nil), z)) -> APP(x'', app(app(app(fold, app(app(fold, f''), x'')), x), z))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(plus, 0), y) -> y


Strategy:

innermost



The Proof could not be continued due to a Timeout.
Innermost Termination of R could not be shown.
Duration:
1:01 minutes