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

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

(0) Obligation:

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

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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(minus(0, Y)) → MARK(0)
ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
ACTIVE(minus(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(geq(X, 0)) → MARK(true)
ACTIVE(geq(0, s(Y))) → MARK(false)
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
ACTIVE(geq(s(X), s(Y))) → GEQ(X, Y)
ACTIVE(div(0, s(Y))) → MARK(0)
ACTIVE(div(s(X), s(Y))) → MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
ACTIVE(div(s(X), s(Y))) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
ACTIVE(div(s(X), s(Y))) → GEQ(X, Y)
ACTIVE(div(s(X), s(Y))) → S(div(minus(X, Y), s(Y)))
ACTIVE(div(s(X), s(Y))) → DIV(minus(X, Y), s(Y))
ACTIVE(div(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(if(true, X, Y)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(div(X1, X2)) → DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
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)
MINUS(mark(X1), X2) → MINUS(X1, X2)
MINUS(X1, mark(X2)) → MINUS(X1, X2)
MINUS(active(X1), X2) → MINUS(X1, X2)
MINUS(X1, active(X2)) → MINUS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
GEQ(mark(X1), X2) → GEQ(X1, X2)
GEQ(X1, mark(X2)) → GEQ(X1, X2)
GEQ(active(X1), X2) → GEQ(X1, X2)
GEQ(X1, active(X2)) → GEQ(X1, X2)
DIV(mark(X1), X2) → DIV(X1, X2)
DIV(X1, mark(X2)) → DIV(X1, X2)
DIV(active(X1), X2) → DIV(X1, X2)
DIV(X1, active(X2)) → DIV(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)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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:

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(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

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)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • IF(X1, mark(X2), X3) → IF(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

  • IF(mark(X1), X2, X3) → IF(X1, X2, X3)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

  • IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

  • IF(active(X1), X2, X3) → IF(X1, X2, X3)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

  • IF(X1, active(X2), X3) → IF(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

  • IF(X1, X2, active(X3)) → IF(X1, X2, X3)
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

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

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • DIV(X1, mark(X2)) → DIV(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • DIV(mark(X1), X2) → DIV(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • DIV(active(X1), X2) → DIV(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • DIV(X1, active(X2)) → DIV(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(14) TRUE

(15) Obligation:

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

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

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

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

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • GEQ(X1, mark(X2)) → GEQ(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • GEQ(mark(X1), X2) → GEQ(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • GEQ(active(X1), X2) → GEQ(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • GEQ(X1, active(X2)) → GEQ(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(19) TRUE

(20) 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(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • S(active(X)) → S(X)
    The graph contains the following edges 1 > 1

  • S(mark(X)) → S(X)
    The graph contains the following edges 1 > 1

(24) TRUE

(25) Obligation:

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

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

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(26) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

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

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

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

(28) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(X1, mark(X2)) → MINUS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • MINUS(mark(X1), X2) → MINUS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • MINUS(active(X1), X2) → MINUS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • MINUS(X1, active(X2)) → MINUS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(29) TRUE

(30) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(div(s(X), s(Y))) → MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
MARK(s(X)) → MARK(X)
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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.


MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(div(x1, x2)) = x1 + x2   
POL(false) = 0   
POL(geq(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(minus(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
mark(false) → active(false)
mark(true) → active(true)
mark(0) → active(0)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
div(active(X1), X2) → div(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(X1, active(X2)) → div(X1, X2)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
mark(div(X1, X2)) → active(div(mark(X1), X2))
active(geq(s(X), s(Y))) → mark(geq(X, Y))
mark(s(X)) → active(s(mark(X)))
mark(minus(X1, X2)) → active(minus(X1, X2))
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(if(false, X, Y)) → mark(Y)
mark(geq(X1, X2)) → active(geq(X1, X2))
active(if(true, X, Y)) → mark(X)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
active(minus(0, Y)) → mark(0)
active(geq(0, s(Y))) → mark(false)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
active(geq(X, 0)) → mark(true)
active(div(0, s(Y))) → mark(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(32) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(div(s(X), s(Y))) → MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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.


ACTIVE(div(s(X), s(Y))) → MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(div(x1, x2)) = x1 + x2   
POL(false) = 0   
POL(geq(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(minus(x1, x2)) = 0   
POL(s(x1)) = 1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
mark(false) → active(false)
mark(true) → active(true)
mark(0) → active(0)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
div(active(X1), X2) → div(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(X1, active(X2)) → div(X1, X2)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
mark(div(X1, X2)) → active(div(mark(X1), X2))
active(geq(s(X), s(Y))) → mark(geq(X, Y))
mark(s(X)) → active(s(mark(X)))
mark(minus(X1, X2)) → active(minus(X1, X2))
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(if(false, X, Y)) → mark(Y)
mark(geq(X1, X2)) → active(geq(X1, X2))
active(if(true, X, Y)) → mark(X)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
active(minus(0, Y)) → mark(0)
active(geq(0, s(Y))) → mark(false)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
active(geq(X, 0)) → mark(true)
active(div(0, s(Y))) → mark(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(34) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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.


MARK(div(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(div(x1, x2)) = 1 + x1   
POL(false) = 0   
POL(geq(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(minus(x1, x2)) = 0   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
mark(false) → active(false)
mark(true) → active(true)
mark(0) → active(0)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
div(active(X1), X2) → div(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(X1, active(X2)) → div(X1, X2)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
mark(div(X1, X2)) → active(div(mark(X1), X2))
active(geq(s(X), s(Y))) → mark(geq(X, Y))
mark(s(X)) → active(s(mark(X)))
mark(minus(X1, X2)) → active(minus(X1, X2))
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(if(false, X, Y)) → mark(Y)
mark(geq(X1, X2)) → active(geq(X1, X2))
active(if(true, X, Y)) → mark(X)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
active(minus(0, Y)) → mark(0)
active(geq(0, s(Y))) → mark(false)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
active(geq(X, 0)) → mark(true)
active(div(0, s(Y))) → mark(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(36) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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.


ACTIVE(if(false, X, Y)) → MARK(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(div(x1, x2)) = 1   
POL(false) = 1   
POL(geq(x1, x2)) = 1   
POL(if(x1, x2, x3)) = x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(minus(x1, x2)) = 0   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
mark(false) → active(false)
mark(true) → active(true)
mark(0) → active(0)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
div(active(X1), X2) → div(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(X1, active(X2)) → div(X1, X2)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
mark(div(X1, X2)) → active(div(mark(X1), X2))
active(geq(s(X), s(Y))) → mark(geq(X, Y))
mark(s(X)) → active(s(mark(X)))
mark(minus(X1, X2)) → active(minus(X1, X2))
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(if(false, X, Y)) → mark(Y)
mark(geq(X1, X2)) → active(geq(X1, X2))
active(if(true, X, Y)) → mark(X)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
active(minus(0, Y)) → mark(0)
active(geq(0, s(Y))) → mark(false)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
active(geq(X, 0)) → mark(true)
active(div(0, s(Y))) → mark(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(38) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(if(true, X, Y)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(div(x1, x2)) = 1   
POL(false) = 0   
POL(geq(x1, x2)) = 0   
POL(if(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(minus(x1, x2)) = 0   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
mark(false) → active(false)
mark(true) → active(true)
mark(0) → active(0)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
div(active(X1), X2) → div(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(X1, active(X2)) → div(X1, X2)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
mark(div(X1, X2)) → active(div(mark(X1), X2))
active(geq(s(X), s(Y))) → mark(geq(X, Y))
mark(s(X)) → active(s(mark(X)))
mark(minus(X1, X2)) → active(minus(X1, X2))
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(if(false, X, Y)) → mark(Y)
mark(geq(X1, X2)) → active(geq(X1, X2))
active(if(true, X, Y)) → mark(X)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
active(minus(0, Y)) → mark(0)
active(geq(0, s(Y))) → mark(false)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
active(geq(X, 0)) → mark(true)
active(div(0, s(Y))) → mark(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(40) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = x1   
POL(active(x1)) = 0   
POL(div(x1, x2)) = 1   
POL(false) = 0   
POL(geq(x1, x2)) = 0   
POL(if(x1, x2, x3)) = x3   
POL(mark(x1)) = 0   
POL(minus(x1, x2)) = 0   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)

(42) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(div(x1, x2)) = 0   
POL(false) = 0   
POL(geq(x1, x2)) = 1   
POL(if(x1, x2, x3)) = 0   
POL(mark(x1)) = 0   
POL(minus(x1, x2)) = 1   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)

(44) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(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)

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

(45) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(46) Obligation:

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

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))

The TRS R consists of the following rules:

geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)

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

(47) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
The following rules are removed from R:

geq(X1, active(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 2 + x1   
POL(MARK(x1)) = 1 + 2·x1   
POL(active(x1)) = 2·x1   
POL(geq(x1, x2)) = 2 + x1 + 2·x2   
POL(mark(x1)) = 2·x1   
POL(minus(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(s(x1)) = 1 + 2·x1   

(48) Obligation:

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE