YES Termination w.r.t. Q proof of AProVE_07_wiehe12.ari

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

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

(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) YES

(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) YES

(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) YES

(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,JAR06].


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 + 1A·x1 + 0A·x2

POL(s(x1)) = 0A + 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)) = 0A + 1A·x1

POL(not(x1)) = 0A + 1A·x1

POL(0) = 2A

POL(true) = 0A

POL(false) = 0A

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(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,JAR06].


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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( P(x1, x2) ) = x1 + 1

POL( if(x1, ..., x3) ) = 2x1 + x2 + x3

POL( gt(x1, x2) ) = max{0, -1}

POL( s(x1) ) = 2x1 + 2

POL( 0 ) = 0

POL( true ) = 0

POL( false ) = 0

POL( id(x1) ) = x1 + 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y

(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,JAR06].


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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( P(x1, x2) ) = 2x1 + 2

POL( if(x1, ..., x3) ) = x1 + x2 + x3

POL( gt(x1, x2) ) = 0

POL( s(x1) ) = 2x1 + 1

POL( 0 ) = 0

POL( true ) = 0

POL( false ) = 0

POL( id(x1) ) = x1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
id(x) → x
if(true, x, y) → x
if(false, x, y) → y

(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) YES

(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,JAR06].


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

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
dummyConstant=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] 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) YES

(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) YES

(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) YES

(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,JAR06].


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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( G(x1, x2) ) = 2x1 + x2

POL( minus(x1, x2) ) = x1

POL( m(x1, x2) ) = x1 + x2

POL( 0 ) = 0

POL( s(x1) ) = 2x1 + 1

POL( n(x1, x2) ) = x1

POL( k(x1, x2) ) = x1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(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) YES