Term Rewriting System R:
[y, x]
int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

INT(0, s(y)) -> INT(s(0), s(y))
INT(s(x), s(y)) -> INTLIST(int(x, y))
INT(s(x), s(y)) -> INT(x, y)
INTLIST(.(x, y)) -> INTLIST(y)

Furthermore, R contains two SCCs.

R
DPs
→DP Problem 1
Polynomial Ordering
→DP Problem 2
Polo

Dependency Pair:

INTLIST(.(x, y)) -> INTLIST(y)

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

INTLIST(.(x, y)) -> INTLIST(y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(INT_LIST(x1)) =  x1 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 3
Dependency Graph
→DP Problem 2
Polo

Dependency Pair:

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polynomial Ordering

Dependency Pairs:

INT(s(x), s(y)) -> INT(x, y)
INT(0, s(y)) -> INT(s(0), s(y))

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(s(x1)) =  1 + x1 POL(INT(x1, x2)) =  x2

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 4
Dependency Graph

Dependency Pair:

INT(0, s(y)) -> INT(s(0), s(y))

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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