Termination w.r.t. Q proof of /tmp/tmp8lpjiU/Liveness6.2.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:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

rest(cons(x, y)) → sent(y)

Used ordering:
Polynomial interpretation [25]:

POL(check(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(nil) = 0
POL(rest(x₁)) = x₁
POL(sent(x₁)) = x₁
POL(top(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Used ordering:
Polynomial interpretation [25]:

POL(check(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(nil) = 0
POL(rest(x₁)) = x₁
POL(sent(x₁)) = 2·x₁
POL(top(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))

Q is empty.

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

TOP(sent(x)) → REST(x)
TOP(sent(x)) → CHECK(rest(x))
TOP(sent(x)) → TOP(check(rest(x)))
CHECK(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(x)
CHECK(rest(x)) → REST(check(x))

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

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

TOP(sent(x)) → REST(x)
TOP(sent(x)) → CHECK(rest(x))
TOP(sent(x)) → TOP(check(rest(x)))
CHECK(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(x)
CHECK(rest(x)) → REST(check(x))

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

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

CHECK(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

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

CHECK(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(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:

CHECK(sent(x)) → CHECK(x)
The graph contains the following edges 1 > 1
CHECK(rest(x)) → CHECK(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

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

TOP(sent(x)) → TOP(check(rest(x)))

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ RFCMatchBoundsDPProof

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

TOP(sent(x)) → TOP(check(rest(x)))

The TRS R consists of the following rules:

rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
Finiteness of the DP problem can be shown by a matchbound of 2.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

TOP(sent(x)) → TOP(check(rest(x)))

To find matches we regarded all rules of R and P:

rest(nil) → sent(nil)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
TOP(sent(x)) → TOP(check(rest(x)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

52, 53, 54, 55, 56, 57, 58, 59, 60, 61

Node 52 is start node and node 53 is final node.

Those nodes are connect through the following edges:

52 to 54 labelled TOP_1(0)
52 to 59 labelled TOP_1(1)
53 to 53 labelled #_1(0)
54 to 55 labelled check_1(0)
54 to 56 labelled rest_1(1)
54 to 58 labelled sent_1(1)
55 to 53 labelled rest_1(0)
55 to 57 labelled sent_1(1)
56 to 53 labelled check_1(1)
56 to 56 labelled sent_1(1), rest_1(1)
57 to 53 labelled nil(1)
58 to 57 labelled check_1(1)
59 to 60 labelled check_1(1)
59 to 61 labelled rest_1(2)
60 to 58 labelled rest_1(1)
61 to 58 labelled check_1(2)