Termination Proof using AProVE

Term Rewriting System R:
[y, x, f, xs, g]
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) -> app(app(g, x), y)
inc -> app(map, app(app(curry, plus), app(s, 0)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(app(curry, g), x), y) -> APP(g, x)
INC -> APP(map, app(app(curry, plus), app(s, 0)))
INC -> APP(app(curry, plus), app(s, 0))
INC -> APP(curry, plus)
INC -> APP(s, 0)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

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

Rules:

app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) -> app(app(g, x), y)
inc -> app(map, app(app(curry, plus), app(s, 0)))

Strategy:

innermost

As we are in the innermost case, we can delete all 6 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳A-Transformation

       →DP Problem 2

         ↳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 3

             ↳ATrans

...

               →DP Problem 4

                 ↳Size-Change Principle

       →DP Problem 2

         ↳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)

Dependency Pairs:

APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)

Rules:

app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(curry, g), x), y) -> app(app(g, x), y)
inc -> app(map, app(app(curry, plus), app(s, 0)))

Strategy:

innermost

As we are in the innermost case, we can delete all 1 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Polynomial Ordering

Dependency Pairs:

Rules:

app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, 0), y) -> y
app(app(map, f), nil) -> nil
app(app(app(curry, g), x), y) -> app(app(g, x), y)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)

Additionally, the following usable rules w.r.t. the implicit AFS can be oriented:

app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, 0), y) -> y
app(app(map, f), nil) -> nil
app(app(app(curry, g), x), y) -> app(app(g, x), y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus) = 0

POL(0) = 0

POL(curry) = 1

POL(cons) = 0

POL(map) = 1

POL(nil) = 0

POL(s) = 0

POL(app(x₁, x₂)) = 1 + x₁ + x₁·x₂

POL(APP(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Polo

...

               →DP Problem 6

                 ↳Usable Rules (Innermost)

Dependency Pair:

APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)

Rules:

app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, 0), y) -> y
app(app(map, f), nil) -> nil
app(app(app(curry, g), x), y) -> app(app(g, x), y)

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Polo

...

               →DP Problem 7

                 ↳A-Transformation

Dependency Pair:

APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)

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 5

             ↳Polo

...

               →DP Problem 8

                 ↳Size-Change Principle

Dependency Pair:

MAP(f, cons(x, xs)) -> MAP(f, xs)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

MAP(f, cons(x, xs)) -> MAP(f, xs)

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:

cons(x₁, x₂) -> cons(x₁, x₂)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes

POL(plus)	= 0
POL(0)	= 0
POL(curry)	= 1
POL(cons)	= 0
POL(map)	= 1
POL(nil)	= 0
POL(s)	= 0
POL(app(x₁, x₂))	= 1 + x₁ + x₁·x₂
POL(APP(x₁, x₂))	= x₁