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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

Q is empty.

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, 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:

+(x, 0) → x
+(0, y) → y
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))
Used ordering:
Polynomial interpretation [25]:

POL(+(x1, x2)) = 1 + x1 + x2   
POL(0) = 1   
POL(f(x1)) = 1 + x1   
POL(g(x1)) = 2 + 2·x1   
POL(h(x1, x2)) = 1 + x1 + 2·x2   
POL(s(x1)) = 2 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))

Q is empty.

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

+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), 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:

+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
Used ordering:
Polynomial interpretation [25]:

POL(+(x1, x2)) = 1 + x1 + 2·x2   
POL(0) = 0   
POL(f(x1)) = 2 + 2·x1   
POL(g(x1)) = 2·x1   
POL(h(x1, x2)) = 2 + 2·x1 + x2   
POL(s(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Q is empty.

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

+(x, s(y)) → s(+(x, y))
+(s(x), y) → 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:

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

POL(+(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(s(x1)) = 1 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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.