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:

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

Q is empty.

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

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

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

f(x, y) → x
Used ordering:
Polynomial interpretation [25]:

POL(a) = 0   
POL(b) = 0   
POL(f(x1, x2)) = 2 + x1 + x2   
POL(g(x1)) = 2·x1   
POL(h(x1, x2, x3)) = x1 + 2·x2 + 2·x3   
POL(i(x1)) = 2 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

Q is empty.

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

g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

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

i(x) → f(x, x)
Used ordering:
Polynomial interpretation [25]:

POL(a) = 0   
POL(b) = 0   
POL(f(x1, x2)) = 1 + x1 + x2   
POL(g(x1)) = x1   
POL(h(x1, x2, x3)) = x1 + x2 + 2·x3   
POL(i(x1)) = 2 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost

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

g(a) → h(a, b, a)
h(x, x, y) → g(x)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

g(a) → h(a, b, a)
h(x, x, y) → g(x)

The signature Sigma is {h, g}

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof

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

g(a) → h(a, b, a)
h(x, x, y) → g(x)

The set Q consists of the following terms:

g(a)
h(x0, 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:

H(x, x, y) → G(x)
G(a) → H(a, b, a)

The TRS R consists of the following rules:

g(a) → h(a, b, a)
h(x, x, y) → g(x)

The set Q consists of the following terms:

g(a)
h(x0, x0, x1)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

H(x, x, y) → G(x)
G(a) → H(a, b, a)

The TRS R consists of the following rules:

g(a) → h(a, b, a)
h(x, x, y) → g(x)

The set Q consists of the following terms:

g(a)
h(x0, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.