Term Rewriting System R:
[YS, X, XS, Y, L]
app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(cons(X, XS), YS) -> APP(XS, YS)
FROM(X) -> FROM(s(X))
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nil))
PREFIX(L) -> PREFIX(L)

Furthermore, R contains four SCCs.

R
DPs
→DP Problem 1
Remaining Obligation(s)
→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:

APP(cons(X, XS), YS) -> APP(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

R
DPs
→DP Problem 1
Remaining Obligation(s)
→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:

APP(cons(X, XS), YS) -> APP(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

R
DPs
→DP Problem 1
Remaining Obligation(s)
→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:

APP(cons(X, XS), YS) -> APP(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

R
DPs
→DP Problem 1
Remaining Obligation(s)
→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:

APP(cons(X, XS), YS) -> APP(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))