Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 UsableRulesProof (⇔)
    ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 QDPOrderProof (⇔)
    ↳26 QDP
    ↳27 PisEmptyProof (⇔)
    ↳28 TRUE
  ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE
  ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 QDPSizeChangeProof (⇔)
    ↳38 TRUE
  ↳39 QDP
    ↳40 UsableRulesProof (⇔)
    ↳41 QDP
    ↳42 QDPSizeChangeProof (⇔)
    ↳43 TRUE
  ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 PisEmptyProof (⇔)
    ↳48 TRUE

(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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

H(s(x)) → F(x)
F(s(x)) → H(x)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • F(s(x)) → H(x)
    The graph contains the following edges 1 > 1

  • H(s(x)) → F(x)
    The graph contains the following edges 1 > 1

(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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


P(s(x), s(y)) → P(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(P(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(s(x1)) = 1A + 2A·x1

POL(if(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 0A·x3

POL(gt(x1, x2)) = -I + 0A·x1 + -I·x2

POL(id(x1)) = -I + 1A·x1

POL(not(x1)) = 0A + 0A·x1

POL(false) = 0A

POL(true) = 0A

POL(0) = 0A

The following usable rules [FROCOS05] were oriented:

if(false, x, y) → y
not(x) → if(x, false, true)
id(x) → x
if(true, x, y) → x
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)

(22) 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(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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


P(id(x), s(y)) → P(x, if(gt(s(y), y), y, s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(P(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/1\
\1/
 +
/11\
\11/
·x1

POL(if(x1, x2, x3)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/10\
\01/
·x2 +
/10\
\01/
·x3

POL(gt(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(id(x1)) =
/0\
\1/
 +
/10\
\01/
·x1

POL(false) =
/1\
\1/

POL(true) =
/1\
\1/

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

if(false, x, y) → y
id(x) → x
if(true, x, y) → x

(24) 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))

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.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


P(s(x), x) → P(if(gt(x, x), id(x), id(x)), s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(P(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/1\
\0/
 +
/11\
\11/
·x1

POL(if(x1, x2, x3)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/10\
\01/
·x2 +
/10\
\01/
·x3

POL(gt(x1, x2)) =
/0\
\0/
 +
/01\
\01/
·x1 +
/00\
\01/
·x2

POL(id(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(false) =
/1\
\1/

POL(true) =
/1\
\1/

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

if(false, x, y) → y
id(x) → x
if(true, x, y) → x

(26) 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.

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) 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.

(30) 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: Matrix interpretation [MATRO]:

POL(K(x1, x2)) =
/0\
\0/
 +
/01\
\00/
·x1 +
/00\
\01/
·x2

POL(s(x1)) =
/0\
\1/
 +
/11\
\11/
·x1

POL(minus(x1, x2)) =
/0\
\0/
 +
/11\
\11/
·x1 +
/11\
\00/
·x2

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(31) 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.

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) 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.

(35) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(36) Obligation:

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

M(s(x), s(y)) → M(x, y)

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

(37) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(38) TRUE

(39) 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.

(40) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(41) Obligation:

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

N(s(x), s(y)) → N(x, y)

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

(42) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(43) TRUE

(44) 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.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(G(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/01\
\00/
·x2

POL(s(x1)) =
/0\
\1/
 +
/11\
\11/
·x1

POL(minus(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2

POL(m(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/10\
\01/
·x2

POL(n(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2

POL(k(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(46) 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.

(47) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE