Termination w.r.t. Q proof of /tmp/tmpt2NBGh/Ex1_GM03

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 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 PisEmptyProof (⇔)

    ↳39 TRUE

  ↳40 QDP

    ↳41 QDPOrderProof (⇔)

    ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 PisEmptyProof (⇔)

    ↳46 TRUE

  ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 PisEmptyProof (⇔)

    ↳53 TRUE

  ↳54 QDP

(0) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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:

ACTIVE(leq(s(X), s(Y))) → LEQ(X, Y)
ACTIVE(diff(X, Y)) → IF(leq(X, Y), 0, s(diff(p(X), Y)))
ACTIVE(diff(X, Y)) → LEQ(X, Y)
ACTIVE(diff(X, Y)) → S(diff(p(X), Y))
ACTIVE(diff(X, Y)) → DIFF(p(X), Y)
ACTIVE(diff(X, Y)) → P(X)
ACTIVE(p(X)) → P(active(X))
ACTIVE(p(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(leq(X1, X2)) → LEQ(active(X1), X2)
ACTIVE(leq(X1, X2)) → ACTIVE(X1)
ACTIVE(leq(X1, X2)) → LEQ(X1, active(X2))
ACTIVE(leq(X1, X2)) → ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(diff(X1, X2)) → DIFF(active(X1), X2)
ACTIVE(diff(X1, X2)) → ACTIVE(X1)
ACTIVE(diff(X1, X2)) → DIFF(X1, active(X2))
ACTIVE(diff(X1, X2)) → ACTIVE(X2)
P(mark(X)) → P(X)
S(mark(X)) → S(X)
LEQ(mark(X1), X2) → LEQ(X1, X2)
LEQ(X1, mark(X2)) → LEQ(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
DIFF(mark(X1), X2) → DIFF(X1, X2)
DIFF(X1, mark(X2)) → DIFF(X1, X2)
PROPER(p(X)) → P(proper(X))
PROPER(p(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(leq(X1, X2)) → LEQ(proper(X1), proper(X2))
PROPER(leq(X1, X2)) → PROPER(X1)
PROPER(leq(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(diff(X1, X2)) → DIFF(proper(X1), proper(X2))
PROPER(diff(X1, X2)) → PROPER(X1)
PROPER(diff(X1, X2)) → PROPER(X2)
P(ok(X)) → P(X)
S(ok(X)) → S(X)
LEQ(ok(X1), ok(X2)) → LEQ(X1, X2)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
DIFF(ok(X1), ok(X2)) → DIFF(X1, X2)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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 20 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

DIFF(X1, mark(X2)) → DIFF(X1, X2)
DIFF(mark(X1), X2) → DIFF(X1, X2)
DIFF(ok(X1), ok(X2)) → DIFF(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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.

DIFF(X1, mark(X2)) → DIFF(X1, X2)
DIFF(ok(X1), ok(X2)) → DIFF(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

ok₁ > DIFF₁

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

DIFF(mark(X1), X2) → DIFF(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIFF(mark(X1), X2) → DIFF(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

mark₁ > DIFF₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = IF(x3)
ok(x1) = ok(x1)
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

mark₁ > IF₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = IF(x1)
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

mark₁ > IF₁

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(mark(X1), X2) → LEQ(X1, X2)
LEQ(ok(X1), ok(X2)) → LEQ(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(ok(X1), ok(X2)) → LEQ(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

ok₁ > LEQ₁

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

LEQ(mark(X1), X2) → LEQ(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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.

LEQ(mark(X1), X2) → LEQ(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

mark₁ > LEQ₁

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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:

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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.

S(ok(X)) → S(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

mark₁ > S₁

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

P(ok(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

P(ok(X)) → P(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

P(mark(X)) → P(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

mark₁ > P₁

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(p(X)) → PROPER(X)
PROPER(leq(X1, X2)) → PROPER(X1)
PROPER(leq(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(diff(X1, X2)) → PROPER(X1)
PROPER(diff(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)
PROPER(leq(X1, X2)) → PROPER(X1)
PROPER(leq(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(diff(X1, X2)) → PROPER(X1)
PROPER(diff(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
s(x1) = s(x1)
p(x1) = x1
leq(x1, x2) = leq(x1, x2)
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

PROPER(p(X)) → PROPER(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(p(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

p₁ > PROPER₁

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(p(X)) → ACTIVE(X)
ACTIVE(leq(X1, X2)) → ACTIVE(X1)
ACTIVE(leq(X1, X2)) → ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(diff(X1, X2)) → ACTIVE(X1)
ACTIVE(diff(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(p(X)) → ACTIVE(X)
ACTIVE(leq(X1, X2)) → ACTIVE(X1)
ACTIVE(leq(X1, X2)) → ACTIVE(X2)
ACTIVE(diff(X1, X2)) → ACTIVE(X1)
ACTIVE(diff(X1, X2)) → ACTIVE(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
s(x1) = s(x1)
p(x1) = p(x1)
leq(x1, x2) = leq(x1, x2)
if(x1, x2, x3) = x1
diff(x1, x2) = diff(x1, x2)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(49) Obligation:

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

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
if(x1, x2, x3) = if(x1)

Lexicographic Path Order [LPO].
Precedence:

if₁ > ACTIVE₁

The following usable rules [FROCOS05] were oriented: none

(51) Obligation:

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE

(54) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) → p(active(X))
active(s(X)) → s(active(X))
active(leq(X1, X2)) → leq(active(X1), X2)
active(leq(X1, X2)) → leq(X1, active(X2))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(diff(X1, X2)) → diff(active(X1), X2)
active(diff(X1, X2)) → diff(X1, active(X2))
p(mark(X)) → mark(p(X))
s(mark(X)) → mark(s(X))
leq(mark(X1), X2) → mark(leq(X1, X2))
leq(X1, mark(X2)) → mark(leq(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
diff(mark(X1), X2) → mark(diff(X1, X2))
diff(X1, mark(X2)) → mark(diff(X1, X2))
proper(p(X)) → p(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(leq(X1, X2)) → leq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) → diff(proper(X1), proper(X2))
p(ok(X)) → ok(p(X))
s(ok(X)) → ok(s(X))
leq(ok(X1), ok(X2)) → ok(leq(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
diff(ok(X1), ok(X2)) → ok(diff(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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