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(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)

Q is empty.

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

f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(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:

g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 1   
POL(f(x1, x2)) = 2 + 2·x1 + x2   
POL(g(x1)) = 2 + x1   
POL(h(x1)) = 1 + 2·x1   
POL(h1(x1, x2)) = 1 + 2·x1 + x2   
POL(h2(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(i(x1)) = 2 + x1   
POL(j(x1, x2)) = 2 + x1 + 2·x2   
POL(k(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))

Q is empty.

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

f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))

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(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(f(x1, x2)) = 2·x1 + 2·x2   
POL(g(x1)) = 1 + x1   
POL(h1(x1, x2)) = 2 + x1 + x2   
POL(h2(x1, x2, x3)) = 1 + 2·x1 + x2 + x3   
POL(j(x1, x2)) = 2 + 2·x1 + x2   
POL(k(x1)) = 1 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RisEmptyProof

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

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