Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

HALF(s(s(x))) → HALF(x)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

HALF(s(s(x))) → HALF(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

HALF(s(s(x))) → HALF(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ 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:

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

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

              ↳ 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:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

half(s(0)) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(LOG(x₁)) = 2·x₁
POL(half(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

              ↳ 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:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))

half(s(s(x))) → s(half(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(LOG(x₁)) = 2·x₁
POL(half(x₁)) = x₁
POL(s(x₁)) = 2 + x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ DependencyGraphProof

              ↳ 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:

half(0) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ 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)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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)

From the DPs we obtained the following set of size-change graphs:

BIN2SS(x, cons(1, xs)) → BIN2SS(s(double(x)), xs)
The graph contains the following edges 2 > 2
BIN2SS(x, cons(0, xs)) → BIN2SS(double(x), xs)
The graph contains the following edges 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

From the DPs we obtained the following set of size-change graphs:

LT(s(x), s(y)) → LT(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

From the DPs we obtained the following set of size-change graphs:

EQ(s(x), s(y)) → EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ 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:

bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
bin2ss(x, nil) → x
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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:

bin2s(nil) → 0
bin2s(cons(x, xs)) → bin2ss(x, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
bin2ss(x, nil) → x
bin2ss(x, cons(1, xs)) → bin2ss(s(double(x)), xs)
bin2ss(x, cons(0, xs)) → bin2ss(double(x), xs)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

S2BIN2(x, cons(xs, ys)) → IF2(eq(x, bin2s(xs)), x, xs, ys)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2(false, x, xs, ys) → S2BIN2(x, ys)
The graph contains the following edges 2 >= 1, 4 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                ↳ UsableRulesProof

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:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                ↳ UsableRulesProof

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:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule S2BIN1(x, y, lists) → IF1(lt(y, log(x)), x, y, lists) at position [0] we obtained the following new rules:

S2BIN1(s(0), y1, y2) → IF1(lt(y1, 0), s(0), y1, y2)
S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(half(s(s(x0)))))), s(s(x0)), y1, y2)
S2BIN1(0, y1, y2) → IF1(lt(y1, 0), 0, y1, y2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                ↳ UsableRulesProof

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

IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists))
S2BIN1(s(0), y1, y2) → IF1(lt(y1, 0), s(0), y1, y2)
S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(half(s(s(x0)))))), s(s(x0)), y1, y2)
S2BIN1(0, y1, y2) → IF1(lt(y1, 0), 0, y1, y2)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                ↳ UsableRulesProof

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

IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists))
S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(half(s(s(x0)))))), s(s(x0)), y1, y2)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(half(s(s(x0)))))), s(s(x0)), y1, y2) at position [0,1,0,0] we obtained the following new rules:

S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(s(half(x0))))), s(s(x0)), y1, y2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                ↳ UsableRulesProof

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

IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists))
S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(s(half(x0))))), s(s(x0)), y1, y2)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF1(true, x, y, lists) → S2BIN1(x, s(y), more(lists)) we obtained the following new rules:

IF1(true, s(s(z0)), z1, z2) → S2BIN1(s(s(z0)), s(z1), more(z2))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                ↳ UsableRulesProof

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

IF1(true, s(s(z0)), z1, z2) → S2BIN1(s(s(z0)), s(z1), more(z2))
S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(s(half(x0))))), s(s(x0)), y1, y2)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule S2BIN1(s(s(x0)), y1, y2) → IF1(lt(y1, s(log(s(half(x0))))), s(s(x0)), y1, y2) we obtained the following new rules:

S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(s(z1), s(log(s(half(z0))))), s(s(z0)), s(z1), y_0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                ↳ UsableRulesProof

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

IF1(true, s(s(z0)), z1, z2) → S2BIN1(s(s(z0)), s(z1), more(z2))
S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(s(z1), s(log(s(half(z0))))), s(s(z0)), s(z1), y_0)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(s(z1), s(log(s(half(z0))))), s(s(z0)), s(z1), y_0) at position [0] we obtained the following new rules:

S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(z1, log(s(half(z0)))), s(s(z0)), s(z1), y_0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Instantiation

                ↳ UsableRulesProof

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

IF1(true, s(s(z0)), z1, z2) → S2BIN1(s(s(z0)), s(z1), more(z2))
S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(z1, log(s(half(z0)))), s(s(z0)), s(z1), y_0)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF1(true, s(s(z0)), z1, z2) → S2BIN1(s(s(z0)), s(z1), more(z2)) we obtained the following new rules:

IF1(true, s(s(z0)), s(z1), z2) → S2BIN1(s(s(z0)), s(s(z1)), more(z2))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                ↳ UsableRulesProof

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

IF1(true, s(s(z0)), s(z1), z2) → S2BIN1(s(s(z0)), s(s(z1)), more(z2))
S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(z1, log(s(half(z0)))), s(s(z0)), s(z1), y_0)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule S2BIN1(s(s(z0)), s(z1), y_0) → IF1(lt(z1, log(s(half(z0)))), s(s(z0)), s(z1), y_0) we obtained the following new rules:

S2BIN1(s(s(z0)), s(s(z1)), y_0) → IF1(lt(s(z1), log(s(half(z0)))), s(s(z0)), s(s(z1)), y_0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                ↳ UsableRulesProof

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

IF1(true, s(s(z0)), s(z1), z2) → S2BIN1(s(s(z0)), s(s(z1)), more(z2))
S2BIN1(s(s(z0)), s(s(z1)), y_0) → IF1(lt(s(z1), log(s(half(z0)))), s(s(z0)), s(s(z1)), y_0)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                          ↳ QDP

                ↳ UsableRulesProof

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

IF1(true, s(s(z0)), s(z1), z2) → S2BIN1(s(s(z0)), s(s(z1)), more(z2))
S2BIN1(s(s(z0)), s(s(z1)), y_0) → IF1(lt(s(z1), log(s(half(z0)))), s(s(z0)), s(s(z1)), y_0)

The TRS R consists of the following rules:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

Q is empty.
We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
bin2s(nil)
bin2s(cons(x0, x1))
bin2ss(x0, nil)
bin2ss(x0, cons(0, x1))
bin2ss(x0, cons(1, x1))
s2bin(x0)
s2bin1(x0, x1, x2)
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
s2bin2(x0, nil)
s2bin2(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ MNOCProof

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:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
log(0)
log(s(0))
log(s(s(x0)))
more(nil)
more(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ 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:

more(nil) → nil
more(cons(xs, ys)) → cons(cons(0, xs), cons(cons(1, xs), cons(xs, ys)))
log(0) → 0
log(s(0)) → 0
log(s(s(x))) → s(log(half(s(s(x)))))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

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