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

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 QDP

    ↳14 QDPOrderProof (⇔)

    ↳15 QDP

    ↳16 PisEmptyProof (⇔)

    ↳17 TRUE

  ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 PisEmptyProof (⇔)

    ↳27 TRUE

  ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

(0) Obligation:

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.

(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:

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)

(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:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 9 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > HALF₁

Status:

HALF₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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.

(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:

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)

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

(13) Obligation:

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)

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
BIN2SS(x1, x2) = BIN2SS(x1, x2)
cons(x1, x2) = cons(x1, x2)
1 = 1
s(x1) = x1
double(x1) = double
0 = 0

Lexicographic path order with status [LPO].
Precedence:

cons₂ > BIN2SS₂
cons₂ > double
0 > BIN2SS₂
0 > double

Status:

cons₂: [1,2]
BIN2SS₂: [1,2]
1: []
double: []
0: []

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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)

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.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) Obligation:

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)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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

Lexicographic path order with status [LPO].
Precedence:

s₁ > LT₁

Status:

s₁: [1]
LT₁: [1]

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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)

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.

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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

Lexicographic path order with status [LPO].
Precedence:

s₁ > EQ₁

Status:

EQ₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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)

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.

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

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

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S2BIN2(x1, x2) = S2BIN2(x2)
cons(x1, x2) = cons(x2)
IF2(x1, x2, x3, x4) = IF2(x1, x4)
eq(x1, x2) = eq
bin2s(x1) = bin2s(x1)
false = false
0 = 0
true = true
bin2ss(x1, x2) = bin2ss(x1)
1 = 1
s(x1) = x1
double(x1) = double(x1)
nil = nil

Lexicographic path order with status [LPO].
Precedence:

cons₁ > IF2₂ > S2BIN2₁ > bin2s₁ > 0 > double₁
cons₁ > eq > false > S2BIN2₁ > bin2s₁ > 0 > double₁
cons₁ > eq > true > double₁
bin2ss₁ > double₁
1 > double₁
nil > double₁

Status:

eq: []
cons₁: [1]
S2BIN2₁: [1]
true: []
bin2s₁: [1]
double₁: [1]
0: []
IF2₂: [1,2]
bin2ss₁: [1]
false: []
1: []
nil: []

The following usable rules [FROCOS05] were oriented:

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

(30) Obligation:

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)

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.

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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)

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