Termination w.r.t. Q proof of /tmp/tmprpMUZu/wiehe12.trs

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

  ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 PisEmptyProof (⇔)

    ↳30 TRUE

  ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 PisEmptyProof (⇔)

    ↳35 TRUE

  ↳36 QDP

(0) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

G(s(x), s(y)) → IF(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
G(s(x), s(y)) → AND(f(s(x)), f(s(y)))
G(s(x), s(y)) → F(s(x))
G(s(x), s(y)) → F(s(y))
G(s(x), s(y)) → T(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0)))))
G(s(x), s(y)) → G(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))
G(s(x), s(y)) → K(minus(m(x, y), n(x, y)), s(s(0)))
G(s(x), s(y)) → MINUS(m(x, y), n(x, y))
G(s(x), s(y)) → M(x, y)
G(s(x), s(y)) → N(x, y)
G(s(x), s(y)) → K(n(s(x), s(y)), s(s(0)))
G(s(x), s(y)) → N(s(x), s(y))
G(s(x), s(y)) → G(minus(m(x, y), n(x, y)), n(s(x), s(y)))
N(s(x), s(y)) → N(x, y)
M(s(x), s(y)) → M(x, y)
K(s(x), s(y)) → K(minus(x, y), s(y))
K(s(x), s(y)) → MINUS(x, y)
T(x) → P(x, x)
P(s(x), s(y)) → P(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
P(s(x), s(y)) → IF(gt(x, y), x, y)
P(s(x), s(y)) → GT(x, y)
P(s(x), s(y)) → IF(not(gt(x, y)), id(x), id(y))
P(s(x), s(y)) → NOT(gt(x, y))
P(s(x), s(y)) → ID(x)
P(s(x), s(y)) → ID(y)
P(s(x), x) → P(if(gt(x, x), id(x), id(x)), s(x))
P(s(x), x) → IF(gt(x, x), id(x), id(x))
P(s(x), x) → GT(x, x)
P(s(x), x) → ID(x)
P(id(x), s(y)) → P(x, if(gt(s(y), y), y, s(y)))
P(id(x), s(y)) → IF(gt(s(y), y), y, s(y))
P(id(x), s(y)) → GT(s(y), y)
MINUS(s(x), s(y)) → MINUS(x, y)
NOT(x) → IF(x, false, true)
F(s(x)) → H(x)
H(s(x)) → F(x)
GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 25 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GT(s(x), s(y)) → GT(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GT(x1, x2) = GT(x2)
s(x1) = s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > GT₁

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

H(s(x)) → F(x)
F(s(x)) → H(x)

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(s(x)) → F(x)
F(s(x)) → H(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
H(x1) = x1
s(x1) = s(x1)
F(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = MINUS(x2)
s(x1) = s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > MINUS₁

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

P(s(x), x) → P(if(gt(x, x), id(x), id(x)), s(x))
P(s(x), s(y)) → P(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
P(id(x), s(y)) → P(x, if(gt(s(y), y), y, s(y)))

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(21) Obligation:

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

K(s(x), s(y)) → K(minus(x, y), s(y))

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

K(s(x), s(y)) → K(minus(x, y), s(y))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
K(x1, x2) = K(x1)
s(x1) = s(x1)
minus(x1, x2) = x1
0 = 0

Lexicographic Path Order [LPO].
Precedence:

K₁ > s₁
0 > s₁

The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

(23) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

M(s(x), s(y)) → M(x, y)

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

M(s(x), s(y)) → M(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
M(x1, x2) = M(x2)
s(x1) = s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > M₁

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

N(s(x), s(y)) → N(x, y)

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

N(s(x), s(y)) → N(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
N(x1, x2) = N(x2)
s(x1) = s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > N₁

The following usable rules [FROCOS05] were oriented: none

(33) Obligation:

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

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

G(s(x), s(y)) → G(minus(m(x, y), n(x, y)), n(s(x), s(y)))
G(s(x), s(y)) → G(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))

The TRS R consists of the following rules:

g(s(x), s(y)) → if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))
n(0, y) → 0
n(x, 0) → 0
n(s(x), s(y)) → s(n(x, y))
m(0, y) → y
m(x, 0) → x
m(s(x), s(y)) → s(m(x, y))
k(0, s(y)) → 0
k(s(x), s(y)) → s(k(minus(x, y), s(y)))
t(x) → p(x, x)
p(s(x), s(y)) → s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
p(s(x), x) → p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y) → y
p(id(x), s(y)) → s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
and(x, false) → false
and(true, true) → true
f(0) → true
f(s(x)) → h(x)
h(0) → false
h(s(x)) → f(x)
gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)

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