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:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.

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

TERM_SUB2(Term_sub2(m, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_sub2(m, s), t) -> TERM_SUB2(m, Concat2(s, t))
TERM_SUB2(Case3(m, xi, n), s) -> FROZEN4(m, Sum_sub2(xi, s), n, s)
TERM_SUB2(Term_inl1(m), s) -> TERM_SUB2(m, s)
TERM_SUB2(Term_var1(x), Cons_usual3(y, m, s)) -> TERM_SUB2(Term_var1(x), s)
CONCAT2(Cons_usual3(x, m, s), t) -> CONCAT2(s, t)
CONCAT2(Concat2(s, t), u) -> CONCAT2(s, Concat2(t, u))
CONCAT2(Cons_usual3(x, m, s), t) -> TERM_SUB2(m, t)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(n, s)
TERM_SUB2(Case3(m, xi, n), s) -> SUM_SUB2(xi, s)
FROZEN4(m, Sum_constant1(Right), n, s) -> TERM_SUB2(n, s)
CONCAT2(Concat2(s, t), u) -> CONCAT2(t, u)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(n, s)
TERM_SUB2(Term_inr1(m), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_constant1(Left), n, s) -> TERM_SUB2(m, s)
SUM_SUB2(xi, Cons_sum3(psi, k, s)) -> SUM_SUB2(xi, s)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(n, s)
SUM_SUB2(xi, Cons_usual3(y, m, s)) -> SUM_SUB2(xi, s)
TERM_SUB2(Term_var1(x), Cons_sum3(xi, k, s)) -> TERM_SUB2(Term_var1(x), s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(m, s)
CONCAT2(Cons_sum3(xi, k, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(m, s)

The TRS R consists of the following rules:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

TERM_SUB2(Term_sub2(m, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_sub2(m, s), t) -> TERM_SUB2(m, Concat2(s, t))
TERM_SUB2(Case3(m, xi, n), s) -> FROZEN4(m, Sum_sub2(xi, s), n, s)
TERM_SUB2(Term_inl1(m), s) -> TERM_SUB2(m, s)
TERM_SUB2(Term_var1(x), Cons_usual3(y, m, s)) -> TERM_SUB2(Term_var1(x), s)
CONCAT2(Cons_usual3(x, m, s), t) -> CONCAT2(s, t)
CONCAT2(Concat2(s, t), u) -> CONCAT2(s, Concat2(t, u))
CONCAT2(Cons_usual3(x, m, s), t) -> TERM_SUB2(m, t)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(n, s)
TERM_SUB2(Case3(m, xi, n), s) -> SUM_SUB2(xi, s)
FROZEN4(m, Sum_constant1(Right), n, s) -> TERM_SUB2(n, s)
CONCAT2(Concat2(s, t), u) -> CONCAT2(t, u)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(n, s)
TERM_SUB2(Term_inr1(m), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_constant1(Left), n, s) -> TERM_SUB2(m, s)
SUM_SUB2(xi, Cons_sum3(psi, k, s)) -> SUM_SUB2(xi, s)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(n, s)
SUM_SUB2(xi, Cons_usual3(y, m, s)) -> SUM_SUB2(xi, s)
TERM_SUB2(Term_var1(x), Cons_sum3(xi, k, s)) -> TERM_SUB2(Term_var1(x), s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(m, s)
CONCAT2(Cons_sum3(xi, k, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(m, s)

The TRS R consists of the following rules:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

SUM_SUB2(xi, Cons_sum3(psi, k, s)) -> SUM_SUB2(xi, s)
SUM_SUB2(xi, Cons_usual3(y, m, s)) -> SUM_SUB2(xi, s)

The TRS R consists of the following rules:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.
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.


SUM_SUB2(xi, Cons_sum3(psi, k, s)) -> SUM_SUB2(xi, s)
SUM_SUB2(xi, Cons_usual3(y, m, s)) -> SUM_SUB2(xi, s)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(Cons_sum3(x1, x2, x3)) = 3 + x3   
POL(Cons_usual3(x1, x2, x3)) = 3 + 2·x3   
POL(SUM_SUB2(x1, x2)) = 3·x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

TERM_SUB2(Term_var1(x), Cons_usual3(y, m, s)) -> TERM_SUB2(Term_var1(x), s)
TERM_SUB2(Term_var1(x), Cons_sum3(xi, k, s)) -> TERM_SUB2(Term_var1(x), s)

The TRS R consists of the following rules:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.
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.


TERM_SUB2(Term_var1(x), Cons_usual3(y, m, s)) -> TERM_SUB2(Term_var1(x), s)
TERM_SUB2(Term_var1(x), Cons_sum3(xi, k, s)) -> TERM_SUB2(Term_var1(x), s)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(Cons_sum3(x1, x2, x3)) = 3 + 2·x3   
POL(Cons_usual3(x1, x2, x3)) = 3 + x3   
POL(TERM_SUB2(x1, x2)) = 3·x2   
POL(Term_var1(x1)) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

TERM_SUB2(Term_sub2(m, s), t) -> CONCAT2(s, t)
TERM_SUB2(Case3(m, xi, n), s) -> FROZEN4(m, Sum_sub2(xi, s), n, s)
TERM_SUB2(Term_inl1(m), s) -> TERM_SUB2(m, s)
TERM_SUB2(Term_sub2(m, s), t) -> TERM_SUB2(m, Concat2(s, t))
CONCAT2(Cons_usual3(x, m, s), t) -> CONCAT2(s, t)
CONCAT2(Concat2(s, t), u) -> CONCAT2(s, Concat2(t, u))
CONCAT2(Cons_usual3(x, m, s), t) -> TERM_SUB2(m, t)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(n, s)
FROZEN4(m, Sum_constant1(Right), n, s) -> TERM_SUB2(n, s)
CONCAT2(Concat2(s, t), u) -> CONCAT2(t, u)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(n, s)
TERM_SUB2(Term_inr1(m), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_constant1(Left), n, s) -> TERM_SUB2(m, s)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(n, s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(m, s)
CONCAT2(Cons_sum3(xi, k, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(m, s)

The TRS R consists of the following rules:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.
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.


TERM_SUB2(Term_sub2(m, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_inl1(m), s) -> TERM_SUB2(m, s)
TERM_SUB2(Term_sub2(m, s), t) -> TERM_SUB2(m, Concat2(s, t))
CONCAT2(Cons_usual3(x, m, s), t) -> CONCAT2(s, t)
CONCAT2(Concat2(s, t), u) -> CONCAT2(s, Concat2(t, u))
CONCAT2(Cons_usual3(x, m, s), t) -> TERM_SUB2(m, t)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(n, s)
FROZEN4(m, Sum_constant1(Right), n, s) -> TERM_SUB2(n, s)
CONCAT2(Concat2(s, t), u) -> CONCAT2(t, u)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(n, s)
TERM_SUB2(Term_inr1(m), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_constant1(Left), n, s) -> TERM_SUB2(m, s)
TERM_SUB2(Term_app2(m, n), s) -> TERM_SUB2(m, s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(n, s)
FROZEN4(m, Sum_term_var1(xi), n, s) -> TERM_SUB2(m, s)
CONCAT2(Cons_sum3(xi, k, s), t) -> CONCAT2(s, t)
TERM_SUB2(Term_pair2(m, n), s) -> TERM_SUB2(m, s)
The remaining pairs can at least be oriented weakly.

TERM_SUB2(Case3(m, xi, n), s) -> FROZEN4(m, Sum_sub2(xi, s), n, s)
Used ordering: Polynomial interpretation [21]:

POL(CONCAT2(x1, x2)) = 1 + 3·x1   
POL(Case3(x1, x2, x3)) = 3 + x1 + x2 + 2·x3   
POL(Concat2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(Cons_sum3(x1, x2, x3)) = 3 + 2·x3   
POL(Cons_usual3(x1, x2, x3)) = 2 + x2 + 2·x3   
POL(FROZEN4(x1, x2, x3, x4)) = x1 + 2·x2 + x3   
POL(Frozen4(x1, x2, x3, x4)) = x2   
POL(Id) = 0   
POL(Left) = 1   
POL(Right) = 0   
POL(Sum_constant1(x1)) = 1   
POL(Sum_sub2(x1, x2)) = 2   
POL(Sum_term_var1(x1)) = 1   
POL(TERM_SUB2(x1, x2)) = 1 + x1   
POL(Term_app2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(Term_inl1(x1)) = 3 + 2·x1   
POL(Term_inr1(x1)) = 3 + 2·x1   
POL(Term_pair2(x1, x2)) = 3 + x1 + x2   
POL(Term_sub2(x1, x2)) = 3 + x1 + 3·x2   
POL(Term_var1(x1)) = 0   

The following usable rules [14] were oriented:

Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof

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

TERM_SUB2(Case3(m, xi, n), s) -> FROZEN4(m, Sum_sub2(xi, s), n, s)

The TRS R consists of the following rules:

Term_sub2(Case3(m, xi, n), s) -> Frozen4(m, Sum_sub2(xi, s), n, s)
Frozen4(m, Sum_constant1(Left), n, s) -> Term_sub2(m, s)
Frozen4(m, Sum_constant1(Right), n, s) -> Term_sub2(n, s)
Frozen4(m, Sum_term_var1(xi), n, s) -> Case3(Term_sub2(m, s), xi, Term_sub2(n, s))
Term_sub2(Term_app2(m, n), s) -> Term_app2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_pair2(m, n), s) -> Term_pair2(Term_sub2(m, s), Term_sub2(n, s))
Term_sub2(Term_inl1(m), s) -> Term_inl1(Term_sub2(m, s))
Term_sub2(Term_inr1(m), s) -> Term_inr1(Term_sub2(m, s))
Term_sub2(Term_var1(x), Id) -> Term_var1(x)
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> m
Term_sub2(Term_var1(x), Cons_usual3(y, m, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_var1(x), Cons_sum3(xi, k, s)) -> Term_sub2(Term_var1(x), s)
Term_sub2(Term_sub2(m, s), t) -> Term_sub2(m, Concat2(s, t))
Sum_sub2(xi, Id) -> Sum_term_var1(xi)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_constant1(k)
Sum_sub2(xi, Cons_sum3(psi, k, s)) -> Sum_sub2(xi, s)
Sum_sub2(xi, Cons_usual3(y, m, s)) -> Sum_sub2(xi, s)
Concat2(Concat2(s, t), u) -> Concat2(s, Concat2(t, u))
Concat2(Cons_usual3(x, m, s), t) -> Cons_usual3(x, Term_sub2(m, t), Concat2(s, t))
Concat2(Cons_sum3(xi, k, s), t) -> Cons_sum3(xi, k, Concat2(s, t))
Concat2(Id, s) -> s

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.