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 DependencyGraphProof (⇔)
    ↳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)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


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

Recursive Path Order [RPO].
Precedence:
[H1, s1]


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

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.

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(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)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


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, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0

Recursive Path Order [RPO].
Precedence:
[K2, s1]


The following usable rules [FROCOS05] were oriented:

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

(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)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


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)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


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.