Termination w.r.t. Q proof of /tmp/tmpTve6Hw/2.31.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:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(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:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(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:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(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(0) = 2
POL(false) = 1
POL(not(x₁)) = 1 + x₁
POL(odd(x₁)) = 2 + 2·x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RisEmptyProof

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

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