Term Rewriting System R:
[X, Y, L]
eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

EQ(s(X), s(Y)) -> EQ(X, Y)
INF(X) -> INF(s(X))
TAKE(s(X), cons(Y, L)) -> TAKE(X, L)
LENGTH(cons(X, L)) -> LENGTH(L)

Furthermore, R contains four SCCs.

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

Dependency Pair:

EQ(s(X), s(Y)) -> EQ(X, Y)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

The following dependency pair can be strictly oriented:

EQ(s(X), s(Y)) -> EQ(X, Y)

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
EQ(x1, x2) -> EQ(x1, x2)
s(x1) -> s(x1)

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

Dependency Pair:

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

The following remains to be proven:
• Dependency Pair:

INF(X) -> INF(s(X))

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

• Dependency Pair:

TAKE(s(X), cons(Y, L)) -> TAKE(X, L)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

• Dependency Pair:

LENGTH(cons(X, L)) -> LENGTH(L)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

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

The following remains to be proven:
• Dependency Pair:

INF(X) -> INF(s(X))

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

• Dependency Pair:

TAKE(s(X), cons(Y, L)) -> TAKE(X, L)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

• Dependency Pair:

LENGTH(cons(X, L)) -> LENGTH(L)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

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

The following remains to be proven:
• Dependency Pair:

INF(X) -> INF(s(X))

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

• Dependency Pair:

TAKE(s(X), cons(Y, L)) -> TAKE(X, L)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

• Dependency Pair:

LENGTH(cons(X, L)) -> LENGTH(L)

Rules:

eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Strategy:

innermost

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