Termination w.r.t. Q proof of /tmp/tmpNDTDBx/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 QDPOrderProof (⇔)

    ↳11 QDP

    ↳12 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 PisEmptyProof (⇔)

    ↳43 TRUE

  ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 PisEmptyProof (⇔)

    ↳48 TRUE

  ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 PisEmptyProof (⇔)

    ↳55 TRUE

  ↳56 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)

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) = x1
active(x1) = active(x1)
p(x1) = p(x1)
0 = 0
s(x1) = x1
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

DIFF₁ > 0
active₁ > leq₂ > true > mark₁ > 0
active₁ > leq₂ > false > mark₁ > 0
active₁ > if₃ > mark₁ > 0
active₁ > diff₂ > p₁ > mark₁ > 0
top > 0

Status:

DIFF₁: [1]
mark₁: [1]
active₁: [1]
p₁: [1]
0: []
leq₂: [1,2]
true: []
false: []
if₃: [2,3,1]
diff₂: [1,2]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(7) Obligation:

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

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.

(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, x2)
mark(x1) = mark(x1)
ok(x1) = x1
active(x1) = active(x1)
p(x1) = p(x1)
0 = 0
s(x1) = s(x1)
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > p₁ > mark₁ > DIFF₂ > false
active₁ > 0 > true > mark₁ > DIFF₂ > false
active₁ > s₁ > mark₁ > DIFF₂ > false
active₁ > leq₂ > mark₁ > DIFF₂ > false
active₁ > if₃ > mark₁ > DIFF₂ > false
active₁ > diff₂ > mark₁ > DIFF₂ > false
top > false

Status:

DIFF₂: [1,2]
mark₁: [1]
active₁: [1]
p₁: [1]
0: []
s₁: [1]
leq₂: [1,2]
true: []
false: []
if₃: [1,2,3]
diff₂: [2,1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(9) Obligation:

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

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.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x2
ok(x1) = ok(x1)
active(x1) = x1
p(x1) = x1
0 = 0
mark(x1) = mark
s(x1) = s(x1)
leq(x1, x2) = x2
true = true
false = false
if(x1, x2, x3) = if(x2)
diff(x1, x2) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > 0 > ok₁ > mark
proper₁ > 0 > true > mark
proper₁ > 0 > false > mark
proper₁ > s₁ > ok₁ > mark
proper₁ > s₁ > false > mark
proper₁ > if₁ > ok₁ > mark
top > mark

Status:

ok₁: [1]
0: []
mark: []
s₁: [1]
true: []
false: []
if₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(11) 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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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.

(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, x3)
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
p(x1) = x1
0 = 0
s(x1) = s(x1)
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

top > active₁ > s₁ > false > mark₁
top > active₁ > leq₂ > false > mark₁
top > active₁ > true > mark₁
top > active₁ > diff₂ > 0 > mark₁
top > active₁ > diff₂ > if₃ > mark₁

Status:

IF₂: [2,1]
mark₁: [1]
active₁: [1]
0: []
s₁: [1]
leq₂: [2,1]
true: []
false: []
if₃: [3,1,2]
diff₂: [2,1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(16) Obligation:

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

IF(ok(X1), ok(X2), ok(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.

(17) 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(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
p(x1) = p(x1)
0 = 0
mark(x1) = x1
s(x1) = s(x1)
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
diff(x1, x2) = diff(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

IF₁ > true
top > active₁ > p₁ > 0 > ok₁ > true
top > active₁ > p₁ > 0 > false > true
top > active₁ > diff₁ > 0 > ok₁ > true
top > active₁ > diff₁ > 0 > false > true
top > active₁ > diff₁ > s₁ > ok₁ > true
top > active₁ > diff₁ > s₁ > false > true
top > active₁ > diff₁ > leq₂ > ok₁ > true
top > active₁ > diff₁ > leq₂ > false > true
top > active₁ > diff₁ > if₂ > ok₁ > true
top > proper₁ > p₁ > 0 > ok₁ > true
top > proper₁ > p₁ > 0 > false > true
top > proper₁ > diff₁ > 0 > ok₁ > true
top > proper₁ > diff₁ > 0 > false > true
top > proper₁ > diff₁ > s₁ > ok₁ > true
top > proper₁ > diff₁ > s₁ > false > true
top > proper₁ > diff₁ > leq₂ > ok₁ > true
top > proper₁ > diff₁ > leq₂ > false > true
top > proper₁ > diff₁ > if₂ > ok₁ > true

Status:

IF₁: [1]
ok₁: [1]
active₁: [1]
p₁: [1]
0: []
s₁: [1]
leq₂: [1,2]
true: []
false: []
if₂: [2,1]
diff₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(18) 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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) 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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x1
ok(x1) = ok(x1)
active(x1) = active(x1)
p(x1) = x1
0 = 0
s(x1) = x1
leq(x1, x2) = leq(x1)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

LEQ₁ > false
top > active₁ > leq₁ > ok₁ > false
top > active₁ > leq₁ > true > false
top > active₁ > if₃ > ok₁ > false
top > active₁ > diff₁ > ok₁ > false
top > active₁ > diff₁ > 0 > true > false
top > proper₁ > leq₁ > ok₁ > false
top > proper₁ > leq₁ > true > false
top > proper₁ > if₃ > ok₁ > false
top > proper₁ > diff₁ > ok₁ > false
top > proper₁ > diff₁ > 0 > true > false

Status:

LEQ₁: [1]
ok₁: [1]
active₁: [1]
0: []
leq₁: [1]
true: []
false: []
if₃: [2,1,3]
diff₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(23) 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)

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) 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) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
p(x1) = p(x1)
0 = 0
s(x1) = s(x1)
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > p₁ > mark₁
proper₁ > 0 > true > mark₁
proper₁ > false > mark₁
proper₁ > diff₂ > s₁ > leq₂ > mark₁
proper₁ > diff₂ > if₃ > mark₁
top > active₁ > p₁ > mark₁
top > active₁ > 0 > true > mark₁
top > active₁ > false > mark₁
top > active₁ > diff₂ > s₁ > leq₂ > mark₁
top > active₁ > diff₂ > if₃ > mark₁

Status:

mark₁: [1]
active₁: [1]
p₁: [1]
0: []
s₁: [1]
leq₂: [2,1]
true: []
false: []
if₃: [1,3,2]
diff₂: [1,2]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(25) Obligation:

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

LEQ(X1, mark(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.

(26) 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)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LEQ(x1, x2) = LEQ(x1, x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
p(x1) = x1
0 = 0
s(x1) = x1
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = proper(x1)
ok(x1) = ok(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > 0 > true > mark₁ > LEQ₂
proper₁ > 0 > false > mark₁ > LEQ₂
proper₁ > 0 > ok₁
proper₁ > leq₂ > true > mark₁ > LEQ₂
proper₁ > leq₂ > false > mark₁ > LEQ₂
proper₁ > leq₂ > ok₁
proper₁ > if₃ > mark₁ > LEQ₂
proper₁ > if₃ > ok₁
proper₁ > diff₂ > mark₁ > LEQ₂
proper₁ > diff₂ > ok₁
top > active₁ > 0 > true > mark₁ > LEQ₂
top > active₁ > 0 > false > mark₁ > LEQ₂
top > active₁ > 0 > ok₁
top > active₁ > leq₂ > true > mark₁ > LEQ₂
top > active₁ > leq₂ > false > mark₁ > LEQ₂
top > active₁ > leq₂ > ok₁
top > active₁ > if₃ > mark₁ > LEQ₂
top > active₁ > if₃ > ok₁
top > active₁ > diff₂ > mark₁ > LEQ₂
top > active₁ > diff₂ > ok₁

Status:

LEQ₂: [1,2]
mark₁: [1]
active₁: [1]
0: []
leq₂: [1,2]
true: []
false: []
if₃: [1,2,3]
diff₂: [2,1]
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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.

(31) 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
active(x1) = active(x1)
p(x1) = p(x1)
0 = 0
s(x1) = s(x1)
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
diff(x1, x2) = x2
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > p₁ > ok₁
active₁ > 0 > ok₁
active₁ > 0 > true
active₁ > 0 > false
active₁ > s₁ > ok₁
active₁ > s₁ > false
active₁ > leq₂ > ok₁
active₁ > leq₂ > true
active₁ > leq₂ > false
active₁ > if₂ > ok₁
proper₁ > p₁ > ok₁
proper₁ > 0 > ok₁
proper₁ > 0 > true
proper₁ > 0 > false
proper₁ > s₁ > ok₁
proper₁ > s₁ > false
proper₁ > leq₂ > ok₁
proper₁ > leq₂ > true
proper₁ > leq₂ > false
proper₁ > if₂ > ok₁

Status:

ok₁: [1]
active₁: [1]
p₁: [1]
0: []
s₁: [1]
leq₂: [2,1]
true: []
false: []
if₂: [2,1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(32) 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.

(33) 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: Combined order from the following AFS and order.
S(x1) = S(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
p(x1) = x1
0 = 0
s(x1) = x1
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

S₁ > top
active₁ > leq₂ > mark₁ > top
active₁ > leq₂ > true > top
active₁ > false > mark₁ > top
active₁ > if₃ > mark₁ > top
active₁ > diff₂ > 0 > mark₁ > top
active₁ > diff₂ > 0 > true > top

Status:

S₁: [1]
mark₁: [1]
active₁: [1]
0: []
leq₂: [2,1]
true: []
false: []
if₃: [2,3,1]
diff₂: [1,2]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(34) 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.

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) 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.

(38) 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
active(x1) = active(x1)
p(x1) = p(x1)
0 = 0
s(x1) = s(x1)
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
diff(x1, x2) = x2
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > p₁ > ok₁
active₁ > 0 > ok₁
active₁ > 0 > true
active₁ > 0 > false
active₁ > s₁ > ok₁
active₁ > s₁ > false
active₁ > leq₂ > ok₁
active₁ > leq₂ > true
active₁ > leq₂ > false
active₁ > if₂ > ok₁
proper₁ > p₁ > ok₁
proper₁ > 0 > ok₁
proper₁ > 0 > true
proper₁ > 0 > false
proper₁ > s₁ > ok₁
proper₁ > s₁ > false
proper₁ > leq₂ > ok₁
proper₁ > leq₂ > true
proper₁ > leq₂ > false
proper₁ > if₂ > ok₁

Status:

ok₁: [1]
active₁: [1]
p₁: [1]
0: []
s₁: [1]
leq₂: [2,1]
true: []
false: []
if₂: [2,1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(39) 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.

(40) 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: Combined order from the following AFS and order.
P(x1) = P(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
p(x1) = x1
0 = 0
s(x1) = x1
leq(x1, x2) = leq(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

P₁ > top
active₁ > leq₂ > mark₁ > top
active₁ > leq₂ > true > top
active₁ > false > mark₁ > top
active₁ > if₃ > mark₁ > top
active₁ > diff₂ > 0 > mark₁ > top
active₁ > diff₂ > 0 > true > top

Status:

P₁: [1]
mark₁: [1]
active₁: [1]
0: []
leq₂: [2,1]
true: []
false: []
if₃: [2,3,1]
diff₂: [1,2]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(41) 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.

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) 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.

(45) 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(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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
s(x1) = s(x1)
p(x1) = p(x1)
leq(x1, x2) = leq(x1, x2)
if(x1, x2, x3) = if(x1, x2, x3)
diff(x1, x2) = diff(x1, x2)
active(x1) = x1
0 = 0
mark(x1) = mark
true = true
false = false
proper(x1) = proper(x1)
ok(x1) = ok(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

PROPER₁ > true
proper₁ > p₁ > 0 > ok₁ > top > true
proper₁ > p₁ > mark > top > true
proper₁ > leq₂ > mark > top > true
proper₁ > leq₂ > false > ok₁ > top > true
proper₁ > diff₂ > s₁ > ok₁ > top > true
proper₁ > diff₂ > if₃ > mark > top > true
proper₁ > diff₂ > if₃ > ok₁ > top > true
proper₁ > diff₂ > 0 > ok₁ > top > true

Status:

PROPER₁: [1]
s₁: [1]
p₁: [1]
leq₂: [1,2]
if₃: [1,2,3]
diff₂: [1,2]
0: []
mark: []
true: []
false: []
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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))

(46) 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.

(47) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE

(49) 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.

(50) 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(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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(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)
active(x1) = active(x1)
0 = 0
mark(x1) = x1
true = true
false = false
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

ACTIVE₁ > diff₂
top > active₁ > s₁ > leq₂ > false > diff₂
top > active₁ > s₁ > leq₂ > ok > diff₂
top > active₁ > if₃ > ok > diff₂
top > active₁ > 0 > true > ok > diff₂
top > active₁ > 0 > false > diff₂
top > proper₁ > s₁ > leq₂ > false > diff₂
top > proper₁ > s₁ > leq₂ > ok > diff₂
top > proper₁ > if₃ > ok > diff₂
top > proper₁ > 0 > true > ok > diff₂
top > proper₁ > 0 > false > diff₂

Status:

ACTIVE₁: [1]
s₁: [1]
leq₂: [2,1]
if₃: [2,3,1]
diff₂: [2,1]
active₁: [1]
0: []
true: []
false: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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))

(51) Obligation:

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

ACTIVE(p(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.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(p(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
p(x1) = p(x1)
active(x1) = active(x1)
0 = 0
mark(x1) = mark
s(x1) = x1
leq(x1, x2) = leq
true = true
false = false
if(x1, x2, x3) = if
diff(x1, x2) = diff
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

ACTIVE₁ > false
top > active₁ > leq > mark > p₁ > false
top > active₁ > leq > true > false
top > active₁ > leq > ok > p₁ > false
top > active₁ > if > mark > p₁ > false
top > active₁ > if > ok > p₁ > false
top > active₁ > diff > 0 > true > false
top > active₁ > diff > mark > p₁ > false
top > active₁ > diff > ok > p₁ > false
top > proper₁ > leq > mark > p₁ > false
top > proper₁ > leq > true > false
top > proper₁ > leq > ok > p₁ > false
top > proper₁ > if > mark > p₁ > false
top > proper₁ > if > ok > p₁ > false
top > proper₁ > diff > 0 > true > false
top > proper₁ > diff > mark > p₁ > false
top > proper₁ > diff > ok > p₁ > false

Status:

ACTIVE₁: [1]
p₁: [1]
active₁: [1]
0: []
mark: []
leq: []
true: []
false: []
if: []
diff: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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))

(53) 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.

(54) PisEmptyProof (EQUIVALENT transformation)

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

(55) TRUE

(56) 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.