Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP

(0) Obligation:

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

app(app(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

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(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

(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(app(fold, f), x), app(app(cons, y), z)) → APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(app(fold, f), x), app(app(cons, y), z)) → APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) → APP(app(app(fold, f), x), z)
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(times, app(s, x)), y) → APP(app(plus, app(app(times, x), y)), y)
APP(app(times, app(s, x)), y) → APP(plus, app(app(times, x), y))
APP(app(times, app(s, x)), y) → APP(app(times, x), y)
APP(app(times, app(s, x)), y) → APP(times, x)
SUM → APP(app(fold, add), 0)
SUM → APP(fold, add)
PROD → APP(app(fold, mul), app(s, 0))
PROD → APP(fold, mul)
PROD → APP(s, 0)

The TRS R consists of the following rules:

app(app(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 10 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(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

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: Combined order from the following AFS and order.
plus1(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
s1: [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(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

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

The TRS R consists of the following rules:

app(app(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

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

(13) 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

times1(s(x), y) → times1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(times, app(s, x)), y) → APP(app(times, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
times1(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
s1: [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(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

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

(17) Obligation:

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

APP(app(app(fold, f), x), app(app(cons, y), z)) → APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) → APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(app(fold, f), x), app(app(cons, y), z)) → APP(app(app(fold, f), x), z)

The TRS R consists of the following rules:

app(app(app(fold, f), x), nil) → x
app(app(app(fold, f), x), app(app(cons, y), z)) → app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
sum → app(app(fold, add), 0)
prod → app(app(fold, mul), app(s, 0))

The set Q consists of the following terms:

app(app(app(fold, x0), x1), nil)
app(app(app(fold, x0), x1), app(app(cons, x2), x3))
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
sum
prod

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