Termination w.r.t. Q proof of /tmp/tmpl2eo0K/zerolist.trs

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:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

Q is empty.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SUB(s(x), s(y)) → SUB(x, y)
ZERO(nil) → ZERO2(0, nil)
ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO(cons(x, xs)) → SUB(x, x)
ZERO2(0, cons(x, xs)) → SUB(x, x)
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), nil) → ZERO(nil)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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:

SUB(s(x), s(y)) → SUB(x, y)
ZERO(nil) → ZERO2(0, nil)
ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO(cons(x, xs)) → SUB(x, x)
ZERO2(0, cons(x, xs)) → SUB(x, x)
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), nil) → ZERO(nil)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

SUB(s(x), s(y)) → SUB(x, y)

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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 [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

SUB(s(x), s(y)) → SUB(x, y)

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

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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.[THIEMANN].

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

SUB(s(x), s(y)) → SUB(x, y)

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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.[THIEMANN].

zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZERO2(0, cons(x, xs)) → ZERO(xs)

The remaining pairs can at least be oriented weakly.

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(ZERO(x₁)) = 1 + x₁
POL(ZERO2(x₁, x₂)) = 1 + x₂
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = 0
POL(sub(x₁, x₂)) = 0

The following usable rules [FROCOS05] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ Induction-Processor

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

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))

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

This DP could be deleted by the Induction-Processor:
ZERO(cons(x', xs')) → ZERO2(sub(x', x'), cons(x', xs'))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(ZERO(x₁)) = 1
POL(ZERO2(x₁, x₂)) = x₁
POL(cons(x₁, x₂)) = 0
POL(s(x₁)) = 1
POL(sub(x₁, x₂)) = 1

At least one of these decreasing rules is always used after the deleted DP:
sub(0, 0) → 0
sub(0, s(x91)) → 0

The following formula is valid:
x':sort[a12].sub'(x' , x' )=true

The transformed set:
sub'(0, 0) → true
sub'(s(x1), s(y'')) → sub'(x1, y'')
sub'(s(x5), 0) → false
sub'(0, s(x9)) → true
sub(0, 0) → 0
sub(s(x1), s(y'')) → sub(x1, y'')
sub(s(x5), 0) → s(x5)
sub(0, s(x9)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a12](0, 0) → true
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a19](x0, x2), equal_sort[a19](x1, x3))
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ Induction-Processor

                              ↳ AND

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                ↳ QTRS

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

ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ Induction-Processor

                              ↳ AND

                                ↳ QDP

                                ↳ QTRS

                                  ↳ QTRSRRRProof

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

sub'(0, 0) → true
sub'(s(x1), s(y'')) → sub'(x1, y'')
sub'(s(x5), 0) → false
sub'(0, s(x9)) → true
sub(0, 0) → 0
sub(s(x1), s(y'')) → sub(x1, y'')
sub(s(x5), 0) → s(x5)
sub(0, s(x9)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a12](0, 0) → true
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a19](x0, x2), equal_sort[a19](x1, x3))
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true

Q is empty.

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

sub'(0, 0) → true
sub'(s(x1), s(y'')) → sub'(x1, y'')
sub'(s(x5), 0) → false
sub'(0, s(x9)) → true
sub(0, 0) → 0
sub(s(x1), s(y'')) → sub(x1, y'')
sub(s(x5), 0) → s(x5)
sub(0, s(x9)) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a12](0, 0) → true
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a19](x0, x2), equal_sort[a19](x1, x3))
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true

Q is empty.
Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(and(x₁, x₂)) = 1 + x₁ + x₂
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(equal_bool(x₁, x₂)) = x₁ + x₂
POL(equal_sort[a12](x₁, x₂)) = 2 + x₁ + x₂
POL(equal_sort[a16](x₁, x₂)) = 1 + x₁ + x₂
POL(equal_sort[a19](x₁, x₂)) = x₁ + x₂
POL(equal_sort[a5](x₁, x₂)) = 1 + x₁ + x₂
POL(false) = 2
POL(isa_false(x₁)) = 3 + x₁
POL(isa_true(x₁)) = x₁
POL(not(x₁)) = 3 + x₁
POL(or(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = x₁
POL(sub(x₁, x₂)) = 1 + x₁ + x₂
POL(sub'(x₁, x₂)) = 3 + x₁ + x₂
POL(true) = 0
POL(witness_sort[a16]) = 0
POL(witness_sort[a19]) = 1
POL(witness_sort[a5]) = 0

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

sub'(0, 0) → true
sub'(s(x5), 0) → false
sub'(0, s(x9)) → true
sub(0, 0) → 0
sub(s(x5), 0) → s(x5)
sub(0, s(x9)) → 0
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a12](0, 0) → true
equal_sort[a19](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a19](x0, x2), equal_sort[a19](x1, x3))
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ Induction-Processor

                              ↳ AND

                                ↳ QDP

                                ↳ QTRS

                                  ↳ QTRSRRRProof

                                    ↳ QTRS

                                      ↳ QTRSRRRProof

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

sub'(s(x1), s(y'')) → sub'(x1, y'')
sub(s(x1), s(y'')) → sub(x1, y'')
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
isa_true(true) → true
isa_true(false) → false
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false

Q is empty.

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

sub'(s(x1), s(y'')) → sub'(x1, y'')
sub(s(x1), s(y'')) → sub(x1, y'')
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
isa_true(true) → true
isa_true(false) → false
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false

Q is empty.
Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(cons(x₁, x₂)) = x₁ + x₂
POL(equal_bool(x₁, x₂)) = 1 + x₁ + x₂
POL(equal_sort[a12](x₁, x₂)) = x₁ + x₂
POL(equal_sort[a19](x₁, x₂)) = x₁ + x₂
POL(false) = 0
POL(isa_true(x₁)) = x₁
POL(s(x₁)) = x₁
POL(sub(x₁, x₂)) = x₁ + x₂
POL(sub'(x₁, x₂)) = x₁ + x₂
POL(true) = 0
POL(witness_sort[a19]) = 0

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ Induction-Processor

                              ↳ AND

                                ↳ QDP

                                ↳ QTRS

                                  ↳ QTRSRRRProof

                                    ↳ QTRS

                                      ↳ QTRSRRRProof

                                        ↳ QTRS

                                          ↳ QTRSRRRProof

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

sub'(s(x1), s(y'')) → sub'(x1, y'')
sub(s(x1), s(y'')) → sub(x1, y'')
isa_true(true) → true
isa_true(false) → false
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false

Q is empty.

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

sub'(s(x1), s(y'')) → sub'(x1, y'')
sub(s(x1), s(y'')) → sub(x1, y'')
isa_true(true) → true
isa_true(false) → false
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false

Q is empty.
Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 2
POL(cons(x₁, x₂)) = 2 + x₁ + x₂
POL(equal_sort[a12](x₁, x₂)) = 1 + x₁ + x₂
POL(equal_sort[a19](x₁, x₂)) = 2 + 2·x₁ + x₂
POL(false) = 1
POL(isa_true(x₁)) = 2 + 2·x₁
POL(s(x₁)) = 1 + 2·x₁
POL(sub(x₁, x₂)) = x₁ + x₂
POL(sub'(x₁, x₂)) = 2·x₁ + 2·x₂
POL(true) = 1
POL(witness_sort[a19]) = 2

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

sub'(s(x1), s(y'')) → sub'(x1, y'')
sub(s(x1), s(y'')) → sub(x1, y'')
isa_true(true) → true
isa_true(false) → false
equal_sort[a12](0, s(x0)) → false
equal_sort[a12](s(x0), 0) → false
equal_sort[a12](s(x0), s(x1)) → equal_sort[a12](x0, x1)
equal_sort[a19](cons(x0, x1), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(x0, x1)) → false

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ Induction-Processor

                              ↳ AND

                                ↳ QDP

                                ↳ QTRS

                                  ↳ QTRSRRRProof

                                    ↳ QTRS

                                      ↳ QTRSRRRProof

                                        ↳ QTRS

                                          ↳ QTRSRRRProof

                                            ↳ QTRS

                                              ↳ RisEmptyProof

                                              ↳ RisEmptyProof

                                              ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.