Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

Q is empty.


QTRS
  ↳ AAECC Innermost

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

inc(0) → s(0)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
inc(s(x)) → s(inc(x))
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))

The TRS R 2 is

fg
fh

The signature Sigma is {f, g, h}

↳ QTRS
  ↳ AAECC Innermost
QTRS
      ↳ DependencyPairsProof

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PLUS(x, y) → EQ(x, 0)
PLUS(x, y) → MINUS(x, s(0))
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → PLUS(y, z)
TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)
TIMESITER(x, y, z) → MINUS(x, s(0))
IFPLUS(false, x, y, z) → PLUS(x, z)
EQ(s(x), s(y)) → EQ(x, y)
TIMESITER(x, y, z) → EQ(x, 0)
INC(s(x)) → INC(x)
PLUS(x, y) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

PLUS(x, y) → EQ(x, 0)
PLUS(x, y) → MINUS(x, s(0))
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → PLUS(y, z)
TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)
TIMESITER(x, y, z) → MINUS(x, s(0))
IFPLUS(false, x, y, z) → PLUS(x, z)
EQ(s(x), s(y)) → EQ(x, y)
TIMESITER(x, y, z) → EQ(x, 0)
INC(s(x)) → INC(x)
PLUS(x, y) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

PLUS(x, y) → MINUS(x, s(0))
PLUS(x, y) → EQ(x, 0)
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → PLUS(y, z)
TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)
TIMESITER(x, y, z) → MINUS(x, s(0))
IFPLUS(false, x, y, z) → PLUS(x, z)
EQ(s(x), s(y)) → EQ(x, y)
TIMESITER(x, y, z) → EQ(x, 0)
INC(s(x)) → INC(x)
PLUS(x, y) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 7 less nodes.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EQ(s(x), s(y)) → EQ(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  EQ(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
s1 > EQ1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
s1 > MINUS1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


INC(s(x)) → INC(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INC(x1)  =  INC(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
s1 > INC1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP
                  ↳ QDP

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

IFPLUS(false, x, y, z) → PLUS(x, z)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP

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

TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
fg
fh

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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