Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, x1), x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(plus, app(f, x)), app(app(sumwith, f), xs))
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(plus, app(f, x))
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

plus1(s(x), y) → plus1(x, y)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

[plus1₂, s₁]

Status:

s₁: [1]
plus1₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, f), xs)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(f, x)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, f), xs)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(f, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

APP₂ > sumwith > app₂
cons > sumwith > app₂

Status:

APP₂: [2,1]
cons: []
sumwith: []
app₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE