Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
HALF(s(s(x))) → HALF(x)
EVEN(s(s(x))) → EVEN(x)
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(false, s(x), y) → TIMES(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))
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.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

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)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

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

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)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP

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

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)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof

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(false, s(x), y) → TIMES(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

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(false, s(x), y) → TIMES(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)))
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPOrderProof

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(false, s(x), y) → TIMES(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.
We use the reduction pair processor [15].


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.

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
Used ordering: Polynomial interpretation [25]:

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

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


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.

IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( half(x1) ) =
/0\
\0/
+
/10\
\10/
·x1

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

M( true ) =
/0\
\0/

M( false ) =
/0\
\0/

M( s(x1) ) =
/0\
\1/
+
/11\
\11/
·x1

M( 0 ) =
/1\
\0/

Tuple symbols:
M( TIMES(x1, x2) ) = 1+
[0,1]
·x1+
[0,1]
·x2

M( IF_TIMES(x1, ..., x3) ) = 0+
[0,0]
·x1+
[0,1]
·x2+
[0,1]
·x3


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ DependencyGraphProof

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.