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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 QDPSizeChangeProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 UsableRulesProof (⇔)
    ↳30 QDP
    ↳31 QReductionProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 QDPOrderProof (⇔)
    ↳36 QDP
    ↳37 DependencyGraphProof (⇔)
    ↳38 TRUE

(0) Obligation:

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

EVEN(s(s(x))) → EVEN(x)
HALF(s(s(x))) → HALF(x)
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

(12) 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:

  • PLUS(s(x), y) → PLUS(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(13) TRUE

(14) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

HALF(s(s(x))) → HALF(x)

R is empty.
The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(18) Obligation:

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

HALF(s(s(x))) → HALF(x)

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

(19) 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:

  • HALF(s(s(x))) → HALF(x)
    The graph contains the following edges 1 > 1

(20) TRUE

(21) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

R is empty.
The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(25) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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

(26) 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:

  • EVEN(s(s(x))) → EVEN(x)
    The graph contains the following edges 1 > 1

(27) TRUE

(28) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

(32) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

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.


IF_TIMES(false, s(x), y) → TIMES(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF_TIMES(x1, x2, x3)) = 1 + x2   
POL(TIMES(x1, x2)) = 1 + x1   
POL(even(x1)) = 0   
POL(false) = 0   
POL(half(x1)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(s(x))) → s(half(x))

(34) Obligation:

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

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.


TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(TIMES(x1, x2)) =
/0\
\0/
 +
/10\
\00/
·x1 +
/11\
\10/
·x2

POL(s(x1)) =
/1\
\0/
 +
/11\
\10/
·x1

POL(IF_TIMES(x1, x2, x3)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/01\
\00/
·x2 +
/11\
\10/
·x3

POL(even(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(true) =
/0\
\0/

POL(half(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(0) =
/0\
\0/

POL(false) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

even(s(s(x))) → even(x)
even(s(0)) → false
half(0) → 0
half(s(s(x))) → s(half(x))
even(0) → true

(36) Obligation:

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

IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

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

(37) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(38) TRUE