Termination Proof using AProVE

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.

     ↳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.

     ↳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.

     ↳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.

     ↳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:

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

D_P: 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.

     ↳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.

     ↳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.

     ↳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:

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

D_P: 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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳Narrowing Transformation

Dependency Pairs:

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:

     ↳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