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
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 TRUE
  ↳27 QDP
    ↳28 QDPOrderProof (⇔)
    ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE
  ↳34 QDP

(0) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

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(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

(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(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)
APP(app(lt, app(s, x)), app(s, y)) → APP(lt, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys))))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(lt, x), y))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(lt, x), y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(lt, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(merge, xs)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(eq, x), y))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(eq, x), y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(eq, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(mult, app(s, x)), y) → APP(app(plus, y), app(app(mult, x), y))
APP(app(mult, app(s, x)), y) → APP(plus, y)
APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)
APP(app(mult, app(s, x)), y) → APP(mult, x)
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)
LIST1 → APP(app(map, app(mult, app(s, app(s, 0)))), hamming)
LIST1 → APP(map, app(mult, app(s, app(s, 0))))
LIST1 → APP(mult, app(s, app(s, 0)))
LIST1 → APP(s, app(s, 0))
LIST1 → APP(s, 0)
LIST1 → HAMMING
LIST2 → APP(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
LIST2 → APP(map, app(mult, app(s, app(s, app(s, 0)))))
LIST2 → APP(mult, app(s, app(s, app(s, 0))))
LIST2 → APP(s, app(s, app(s, 0)))
LIST2 → APP(s, app(s, 0))
LIST2 → APP(s, 0)
LIST2 → HAMMING
LIST3 → APP(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
LIST3 → APP(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0)))))))
LIST3 → APP(mult, app(s, app(s, app(s, app(s, app(s, 0))))))
LIST3 → APP(s, app(s, app(s, app(s, app(s, 0)))))
LIST3 → APP(s, app(s, app(s, app(s, 0))))
LIST3 → APP(s, app(s, app(s, 0)))
LIST3 → APP(s, app(s, 0))
LIST3 → APP(s, 0)
LIST3 → HAMMING
HAMMING → APP(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))
HAMMING → APP(cons, app(s, 0))
HAMMING → APP(s, 0)
HAMMING → APP(app(merge, list1), app(app(merge, list2), list3))
HAMMING → APP(merge, list1)
HAMMING → LIST1
HAMMING → APP(app(merge, list2), list3)
HAMMING → APP(merge, list2)
HAMMING → LIST2
HAMMING → LIST3

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 48 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(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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)  =  plus1(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

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(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

mult1(s(x), y) → mult1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
mult1(x1, x2)  =  mult1(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

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(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(18) 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

lt1(s(x), s(y)) → lt1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
lt1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(23) 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

merge1(cons(x, xs), cons(y, ys)) → merge1(xs, ys)
merge1(cons(x, xs), cons(y, ys)) → merge1(xs, cons(y, ys))
merge1(cons(x, xs), cons(y, ys)) → merge1(cons(x, xs), ys)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
cons2 > merge12

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons

Lexicographic Path Order [LPO].
Precedence:
map > APP1 > app2
cons > APP1 > app2

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(30) 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

map1(f, cons(x, xs)) → map1(f, xs)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
map1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

LIST1 → HAMMING
HAMMING → LIST1
HAMMING → LIST2
LIST2 → HAMMING
HAMMING → LIST3
LIST3 → HAMMING

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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