Termination w.r.t. Q proof of /tmp/tmpWEK65o/thiemann21.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

S2BIN2(x, cons(xs, ys)) → EQ(x, bin2s(xs))
LOG(s(s(x))) → HALF(s(s(x)))
BIN2S(cons(x, xs)) → BIN2SS(x, xs)
HALF(s(s(x))) → HALF(x)
S2BIN2(x, cons(xs, ys)) → IF2(eq(x, bin2s(xs)), x, xs, ys)
S2BIN(x) → S2BIN1(x, 0, cons(nil, nil))
S2BIN1(x, y, lists) → IF1(lt(y, log(x)), x, y, lists)
IF1(false, x, y, lists) → S2BIN2(x, lists)
S2BIN2(x, cons(xs, ys)) → BIN2S(xs)
IF2(false, x, xs, ys) → S2BIN2(x, ys)
BIN2SS(x, cons(1, xs)) → BIN2SS(s(double(x)), xs)
IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists))
LT(s(x), s(y)) → LT(x, y)
LOG(s(s(x))) → LOG(half(s(s(x))))
S2BIN1(x, y, lists) → LT(y, log(x))
IF1(true, x, y, lists) → MORE(lists)
S2BIN1(x, y, lists) → LOG(x)
EQ(s(x), s(y)) → EQ(x, y)
BIN2SS(x, cons(0, xs)) → BIN2SS(double(x), xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

S2BIN2(x, cons(xs, ys)) → EQ(x, bin2s(xs))
LOG(s(s(x))) → HALF(s(s(x)))
BIN2S(cons(x, xs)) → BIN2SS(x, xs)
HALF(s(s(x))) → HALF(x)
S2BIN2(x, cons(xs, ys)) → IF2(eq(x, bin2s(xs)), x, xs, ys)
S2BIN(x) → S2BIN1(x, 0, cons(nil, nil))
S2BIN1(x, y, lists) → IF1(lt(y, log(x)), x, y, lists)
IF1(false, x, y, lists) → S2BIN2(x, lists)
S2BIN2(x, cons(xs, ys)) → BIN2S(xs)
IF2(false, x, xs, ys) → S2BIN2(x, ys)
BIN2SS(x, cons(1, xs)) → BIN2SS(s(double(x)), xs)
IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists))
LT(s(x), s(y)) → LT(x, y)
LOG(s(s(x))) → LOG(half(s(s(x))))
S2BIN1(x, y, lists) → LT(y, log(x))
IF1(true, x, y, lists) → MORE(lists)
S2BIN1(x, y, lists) → LOG(x)
EQ(s(x), s(y)) → EQ(x, y)
BIN2SS(x, cons(0, xs)) → BIN2SS(double(x), xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 7 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

HALF(s(s(x))) → HALF(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x₁)) = (2)x_1
POL(s(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LOG(s(s(x))) → LOG(half(s(s(x))))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LOG(s(s(x))) → LOG(half(s(s(x))))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(half(x₁)) = (1/4)x_1
POL(LOG(x₁)) = (4)x_1
POL(s(x₁)) = 7/4 + (4)x_1
POL(0) = 0

The value of delta used in the strict ordering is 105/4.
The following usable rules [17] were oriented:

half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

BIN2SS(x, cons(1, xs)) → BIN2SS(s(double(x)), xs)
BIN2SS(x, cons(0, xs)) → BIN2SS(double(x), xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

BIN2SS(x, cons(1, xs)) → BIN2SS(s(double(x)), xs)
BIN2SS(x, cons(0, xs)) → BIN2SS(double(x), xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 2 + (4)x_2
POL(BIN2SS(x₁, x₂)) = (3)x_1 + (4)x_2
POL(s(x₁)) = 0
POL(double(x₁)) = 0
POL(1) = 3
POL(0) = 0

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LT(s(x), s(y)) → LT(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 4 + (2)x_1
POL(LT(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

EQ(s(x), s(y)) → EQ(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(EQ(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

IF2(false, x, xs, ys) → S2BIN2(x, ys)
S2BIN2(x, cons(xs, ys)) → IF2(eq(x, bin2s(xs)), x, xs, ys)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

IF2(false, x, xs, ys) → S2BIN2(x, ys)
S2BIN2(x, cons(xs, ys)) → IF2(eq(x, bin2s(xs)), x, xs, ys)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(eq(x₁, x₂)) = 0
POL(bin2ss(x₁, x₂)) = 3 + x_1 + x_2
POL(true) = 0
POL(bin2s(x₁)) = 1 + x_1
POL(double(x₁)) = 1
POL(0) = 1
POL(cons(x₁, x₂)) = 3 + x_1 + (2)x_2
POL(S2BIN2(x₁, x₂)) = (2)x_2
POL(IF2(x₁, x₂, x₃, x₄)) = 1 + (2)x_1 + (2)x_3 + (3)x_4
POL(false) = 0
POL(s(x₁)) = 1 + (4)x_1
POL(1) = 3
POL(nil) = 4

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

eq(0, s(y)) → false
eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
eq(s(x), 0) → false

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists))
S2BIN1(x, y, lists) → IF1(lt(y, log(x)), x, y, lists)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
bin2ss(x, nil) → x
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
s2bin(x) → s2bin1(x, 0, cons(nil, nil))
s2bin1(x, y, lists) → if1(lt(y, log(x)), x, y, lists)
if1(true, x, y, lists) → s2bin1(x, s(y), more(lists))
if1(false, x, y, lists) → s2bin2(x, lists)
s2bin2(x, nil) → bug_list_not
s2bin2(x, cons(xs, ys)) → if2(eq(x, bin2s(xs)), x, xs, ys)
if2(true, x, xs, ys) → xs
if2(false, x, xs, ys) → s2bin2(x, ys)

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