Term Rewriting System R:
[k, l, x, y]
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(cons(x, l), k) -> APP(l, k)
SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))
SUM(cons(x, cons(y, l))) -> PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))
SUM(app(l, cons(x, cons(y, k)))) -> APP(l, sum(cons(x, cons(y, k))))
SUM(app(l, cons(x, cons(y, k)))) -> SUM(cons(x, cons(y, k)))
SUM(plus(cons(0, x), cons(y, l))) -> PRED(sum(cons(s(x), cons(y, l))))
SUM(plus(cons(0, x), cons(y, l))) -> SUM(cons(s(x), cons(y, l)))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains four SCCs.

R
DPs
→DP Problem 1
Argument Filtering and Ordering
→DP Problem 2
AFS
→DP Problem 3
AFS
→DP Problem 4
Remaining

Dependency Pair:

APP(cons(x, l), k) -> APP(l, k)

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(cons(x, l), k) -> APP(l, k)

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

R
DPs
→DP Problem 1
AFS
→DP Problem 5
Dependency Graph
→DP Problem 2
AFS
→DP Problem 3
AFS
→DP Problem 4
Remaining

Dependency Pair:

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
AFS
→DP Problem 2
Argument Filtering and Ordering
→DP Problem 3
AFS
→DP Problem 4
Remaining

Dependency Pair:

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

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

The following dependency pair can be strictly oriented:

PLUS(s(x), y) -> PLUS(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:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)

R
DPs
→DP Problem 1
AFS
→DP Problem 2
AFS
→DP Problem 6
Dependency Graph
→DP Problem 3
AFS
→DP Problem 4
Remaining

Dependency Pair:

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
AFS
→DP Problem 2
AFS
→DP Problem 3
Argument Filtering and Ordering
→DP Problem 4
Remaining

Dependency Pair:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

The following dependency pair can be strictly oriented:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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

Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
SUM(x1) -> SUM(x1)
cons(x1, x2) -> cons(x1, x2)
plus(x1, x2) -> x2
s(x1) -> x1

R
DPs
→DP Problem 1
AFS
→DP Problem 2
AFS
→DP Problem 3
AFS
→DP Problem 7
Dependency Graph
→DP Problem 4
Remaining

Dependency Pair:

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
AFS
→DP Problem 2
AFS
→DP Problem 3
AFS
→DP Problem 4
Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

Rules:

app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
sum(plus(cons(0, x), cons(y, l))) -> pred(sum(cons(s(x), cons(y, l))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
pred(cons(s(x), nil)) -> cons(x, nil)

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes