Termination Proof using AProVE

Term Rewriting System R:
[x, y, z, l, l1, l2]
+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(s(x), s(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(s(x), s(y)) -> +'(*(x, y), +(x, y))
*'(s(x), s(y)) -> *'(x, y)
*'(s(x), s(y)) -> +'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
APP(cons(x, l1), l2) -> APP(l1, l2)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
SUM(app(l1, l2)) -> +'(sum(l1), sum(l2))
SUM(app(l1, l2)) -> SUM(l1)
SUM(app(l1, l2)) -> SUM(l2)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)
PROD(app(l1, l2)) -> *'(prod(l1), prod(l2))
PROD(app(l1, l2)) -> PROD(l1)
PROD(app(l1, l2)) -> PROD(l2)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(s(x), s(y)) -> +'(x, y)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(s(x), s(y)) -> +'(x, y)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

+'(x₁, x₂) -> x₁
s(x₁) -> s(x₁)
+(x₁, x₂) -> +(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

APP(cons(x, l1), l2) -> APP(l1, l2)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(cons(x, l1), l2) -> APP(l1, l2)

APP(x₁, x₂) -> APP(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(s(x), s(y)) -> *'(x, y)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

*'(*(x, y), z) -> *'(y, z)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(s(x), s(y)) -> *'(x, y)

*'(x₁, x₂) -> x₁
s(x₁) -> s(x₁)
*(x₁, x₂) -> *(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳AFS

Dependency Pairs:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)

SUM(x₁) -> SUM(x₁)
app(x₁, x₂) -> app(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Argument Filtering and Ordering

Dependency Pairs:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)

PROD(x₁) -> PROD(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
app(x₁, x₂) -> app(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 10

             ↳Dependency Graph

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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