Termination w.r.t. Q proof of /tmp/tmpzogMjH/ExIntrod_Zan97

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 QDPOrderProof (⇔)

    ↳13 QDP

    ↳14 PisEmptyProof (⇔)

    ↳15 TRUE

  ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 QDPOrderProof (⇔)

    ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 QDPOrderProof (⇔)

    ↳58 QDP

    ↳59 QDPOrderProof (⇔)

    ↳60 QDP

    ↳61 PisEmptyProof (⇔)

    ↳62 TRUE

  ↳63 QDP

    ↳64 QDPOrderProof (⇔)

    ↳65 QDP

    ↳66 QDPOrderProof (⇔)

    ↳67 QDP

    ↳68 PisEmptyProof (⇔)

    ↳69 TRUE

  ↳70 QDP

    ↳71 QDPOrderProof (⇔)

    ↳72 QDP

    ↳73 QDPOrderProof (⇔)

    ↳74 QDP

(0) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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(fact(X)) → MARK(if(zero(X), s(0), prod(X, fact(p(X)))))
ACTIVE(fact(X)) → IF(zero(X), s(0), prod(X, fact(p(X))))
ACTIVE(fact(X)) → ZERO(X)
ACTIVE(fact(X)) → S(0)
ACTIVE(fact(X)) → PROD(X, fact(p(X)))
ACTIVE(fact(X)) → FACT(p(X))
ACTIVE(fact(X)) → P(X)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(prod(0, X)) → MARK(0)
ACTIVE(prod(s(X), Y)) → MARK(add(Y, prod(X, Y)))
ACTIVE(prod(s(X), Y)) → ADD(Y, prod(X, Y))
ACTIVE(prod(s(X), Y)) → PROD(X, Y)
ACTIVE(if(true, X, Y)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
ACTIVE(zero(0)) → MARK(true)
ACTIVE(zero(s(X))) → MARK(false)
ACTIVE(p(s(X))) → MARK(X)
MARK(fact(X)) → ACTIVE(fact(mark(X)))
MARK(fact(X)) → FACT(mark(X))
MARK(fact(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(zero(X)) → ACTIVE(zero(mark(X)))
MARK(zero(X)) → ZERO(mark(X))
MARK(zero(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(prod(X1, X2)) → ACTIVE(prod(mark(X1), mark(X2)))
MARK(prod(X1, X2)) → PROD(mark(X1), mark(X2))
MARK(prod(X1, X2)) → MARK(X1)
MARK(prod(X1, X2)) → MARK(X2)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(p(X)) → P(mark(X))
MARK(p(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
FACT(mark(X)) → FACT(X)
FACT(active(X)) → FACT(X)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
ZERO(mark(X)) → ZERO(X)
ZERO(active(X)) → ZERO(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
PROD(mark(X1), X2) → PROD(X1, X2)
PROD(X1, mark(X2)) → PROD(X1, X2)
PROD(active(X1), X2) → PROD(X1, X2)
PROD(X1, active(X2)) → PROD(X1, X2)
P(mark(X)) → P(X)
P(active(X)) → P(X)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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 23 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.

ADD(X1, mark(X2)) → ADD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.

ADD(X1, active(X2)) → ADD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.

ADD(active(X1), X2) → ADD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(mark(X1), X2) → ADD(X1, X2)

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

[ADD₂, mark₁]

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

P(active(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.

P(active(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)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

[P₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(18) 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(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(19) 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) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

PROD(X1, mark(X2)) → PROD(X1, X2)
PROD(mark(X1), X2) → PROD(X1, X2)
PROD(active(X1), X2) → PROD(X1, X2)
PROD(X1, active(X2)) → PROD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.

PROD(X1, mark(X2)) → PROD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

PROD(mark(X1), X2) → PROD(X1, X2)
PROD(active(X1), X2) → PROD(X1, X2)
PROD(X1, active(X2)) → PROD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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.

PROD(X1, active(X2)) → PROD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

PROD(mark(X1), X2) → PROD(X1, X2)
PROD(active(X1), X2) → PROD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROD(active(X1), X2) → PROD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

PROD(mark(X1), X2) → PROD(X1, X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROD(mark(X1), X2) → PROD(X1, X2)

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

[PROD₂, mark₁]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(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)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

[S₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(36) 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(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(37) 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) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

ZERO(active(X)) → ZERO(X)
ZERO(mark(X)) → ZERO(X)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZERO(active(X)) → ZERO(X)

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

Lexicographic Path Order [LPO].
Precedence:

[ZERO₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

ZERO(mark(X)) → ZERO(X)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZERO(mark(X)) → ZERO(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

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

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, X2, mark(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)
mark(x1) = mark(x1)
active(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

[IF₁, mark₁]

The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

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

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, X2, active(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) = x3
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(52) Obligation:

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

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

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, active(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(x2, x3)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(54) Obligation:

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

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

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, mark(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(x2, x3)
mark(x1) = mark(x1)
active(x1) = active

Lexicographic Path Order [LPO].
Precedence:

[IF₂, mark₁]

The following usable rules [FROCOS05] were oriented: none

(56) Obligation:

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

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

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(57) 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)
active(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

[IF₁, mark₁]

The following usable rules [FROCOS05] were oriented: none

(58) Obligation:

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

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

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(active(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) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(60) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(61) PisEmptyProof (EQUIVALENT transformation)

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

(62) TRUE

(63) Obligation:

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

FACT(active(X)) → FACT(X)
FACT(mark(X)) → FACT(X)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FACT(active(X)) → FACT(X)

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

Lexicographic Path Order [LPO].
Precedence:

[FACT₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(65) Obligation:

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

FACT(mark(X)) → FACT(X)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(66) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FACT(mark(X)) → FACT(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(67) Obligation:

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

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(68) PisEmptyProof (EQUIVALENT transformation)

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

(69) TRUE

(70) Obligation:

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

MARK(fact(X)) → ACTIVE(fact(mark(X)))
ACTIVE(fact(X)) → MARK(if(zero(X), s(0), prod(X, fact(p(X)))))
MARK(fact(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
ACTIVE(add(0, X)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(zero(X)) → ACTIVE(zero(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(zero(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(prod(s(X), Y)) → MARK(add(Y, prod(X, Y)))
MARK(s(X)) → MARK(X)
MARK(prod(X1, X2)) → ACTIVE(prod(mark(X1), mark(X2)))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(prod(X1, X2)) → MARK(X1)
MARK(prod(X1, X2)) → MARK(X2)
MARK(p(X)) → ACTIVE(p(mark(X)))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(p(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(p(s(X))) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(71) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK
fact(x1) = fact
ACTIVE(x1) = x1
mark(x1) = mark(x1)
if(x1, x2, x3) = if
zero(x1) = zero
s(x1) = s
0 = 0
prod(x1, x2) = prod
p(x1) = p
add(x1, x2) = add
true = true
false = false
active(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

0 > mark₁ > [MARK, fact, if, zero, prod, p, add] > s > false
0 > mark₁ > [MARK, fact, if, zero, prod, p, add] > true

The following usable rules [FROCOS05] were oriented:

fact(active(X)) → fact(X)
fact(mark(X)) → fact(X)
zero(active(X)) → zero(X)
zero(mark(X)) → zero(X)
p(active(X)) → p(X)
p(mark(X)) → p(X)
prod(X1, mark(X2)) → prod(X1, X2)
prod(mark(X1), X2) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(72) Obligation:

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

MARK(fact(X)) → ACTIVE(fact(mark(X)))
ACTIVE(fact(X)) → MARK(if(zero(X), s(0), prod(X, fact(p(X)))))
MARK(fact(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
ACTIVE(add(0, X)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(zero(X)) → ACTIVE(zero(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(zero(X)) → MARK(X)
ACTIVE(prod(s(X), Y)) → MARK(add(Y, prod(X, Y)))
MARK(s(X)) → MARK(X)
MARK(prod(X1, X2)) → ACTIVE(prod(mark(X1), mark(X2)))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(prod(X1, X2)) → MARK(X1)
MARK(prod(X1, X2)) → MARK(X2)
MARK(p(X)) → ACTIVE(p(mark(X)))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(p(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(p(s(X))) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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

(73) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(zero(X)) → ACTIVE(zero(mark(X)))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK
fact(x1) = fact
ACTIVE(x1) = x1
mark(x1) = mark
if(x1, x2, x3) = if
zero(x1) = zero
s(x1) = s
0 = 0
prod(x1, x2) = prod
p(x1) = p
add(x1, x2) = add
true = true
false = false
active(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

s > [MARK, fact, if, prod, p, add] > [mark, zero, 0]
true > [MARK, fact, if, prod, p, add] > [mark, zero, 0]
false > [MARK, fact, if, prod, p, add] > [mark, zero, 0]

The following usable rules [FROCOS05] were oriented:

fact(active(X)) → fact(X)
fact(mark(X)) → fact(X)
zero(active(X)) → zero(X)
zero(mark(X)) → zero(X)
p(active(X)) → p(X)
p(mark(X)) → p(X)
prod(X1, mark(X2)) → prod(X1, X2)
prod(mark(X1), X2) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

(74) Obligation:

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

MARK(fact(X)) → ACTIVE(fact(mark(X)))
ACTIVE(fact(X)) → MARK(if(zero(X), s(0), prod(X, fact(p(X)))))
MARK(fact(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
ACTIVE(add(0, X)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(zero(X)) → MARK(X)
ACTIVE(prod(s(X), Y)) → MARK(add(Y, prod(X, Y)))
MARK(s(X)) → MARK(X)
MARK(prod(X1, X2)) → ACTIVE(prod(mark(X1), mark(X2)))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(prod(X1, X2)) → MARK(X1)
MARK(prod(X1, X2)) → MARK(X2)
MARK(p(X)) → ACTIVE(p(mark(X)))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(p(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(p(s(X))) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(fact(X)) → mark(if(zero(X), s(0), prod(X, fact(p(X)))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(prod(0, X)) → mark(0)
active(prod(s(X), Y)) → mark(add(Y, prod(X, Y)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(zero(0)) → mark(true)
active(zero(s(X))) → mark(false)
active(p(s(X))) → mark(X)
mark(fact(X)) → active(fact(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(zero(X)) → active(zero(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(prod(X1, X2)) → active(prod(mark(X1), mark(X2)))
mark(p(X)) → active(p(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
fact(mark(X)) → fact(X)
fact(active(X)) → fact(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
zero(mark(X)) → zero(X)
zero(active(X)) → zero(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
prod(mark(X1), X2) → prod(X1, X2)
prod(X1, mark(X2)) → prod(X1, X2)
prod(active(X1), X2) → prod(X1, X2)
prod(X1, active(X2)) → prod(X1, X2)
p(mark(X)) → p(X)
p(active(X)) → p(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)

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