Termination w.r.t. Q proof of /tmp/tmpS5t7_h/Ex1_GM03_iGM.trs

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 QDP

  ↳7 QDP

  ↳8 QDP

  ↳9 QDP

  ↳10 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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(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(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(leq(s(X), s(Y))) → 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(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)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(p(X)) → P(mark(X))
MARK(p(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
MARK(leq(X1, X2)) → LEQ(mark(X1), mark(X2))
MARK(leq(X1, X2)) → MARK(X1)
MARK(leq(X1, X2)) → MARK(X2)
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
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(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
MARK(diff(X1, X2)) → DIFF(mark(X1), mark(X2))
MARK(diff(X1, X2)) → MARK(X1)
MARK(diff(X1, X2)) → MARK(X2)
P(mark(X)) → P(X)
P(active(X)) → P(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
LEQ(mark(X1), X2) → LEQ(X1, X2)
LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(active(X1), X2) → LEQ(X1, X2)
LEQ(X1, active(X2)) → LEQ(X1, X2)
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)
DIFF(mark(X1), X2) → DIFF(X1, X2)
DIFF(X1, mark(X2)) → DIFF(X1, X2)
DIFF(active(X1), X2) → DIFF(X1, X2)
DIFF(X1, active(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(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 6 SCCs with 17 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(active(X1), X2) → DIFF(X1, X2)
DIFF(X1, active(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

(6) 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(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

(7) 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(active(X1), X2) → LEQ(X1, X2)
LEQ(X1, active(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

(8) 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(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

(9) 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(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

(10) Obligation:

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

ACTIVE(p(s(X))) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
ACTIVE(leq(s(X), s(Y))) → MARK(leq(X, Y))
MARK(p(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(leq(X1, X2)) → MARK(X1)
MARK(leq(X1, X2)) → MARK(X2)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
ACTIVE(diff(X, Y)) → MARK(if(leq(X, Y), 0, s(diff(p(X), Y))))
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
MARK(diff(X1, X2)) → MARK(X1)
MARK(diff(X1, X2)) → MARK(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))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
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)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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