Termination w.r.t. Q proof of /tmp/tmpeTd8I2/t000.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:

+(0, 0) → 0
+(0, 1) → 1
+(0, 2) → 2
+(0, 3) → 3
+(0, 4) → 4
+(0, 5) → 5
+(0, 6) → 6
+(0, 7) → 7
+(0, 8) → 8
+(0, 9) → 9
+(1, 0) → 1
+(1, 1) → 2
+(1, 2) → 3
+(1, 3) → 4
+(1, 4) → 5
+(1, 5) → 6
+(1, 6) → 7
+(1, 7) → 8
+(1, 8) → 9
+(1, 9) → c(1, 0)
+(2, 0) → 2
+(2, 1) → 3
+(2, 2) → 4
+(2, 3) → 5
+(2, 4) → 6
+(2, 5) → 7
+(2, 6) → 8
+(2, 7) → 9
+(2, 8) → c(1, 0)
+(2, 9) → c(1, 1)
+(3, 0) → 3
+(3, 1) → 4
+(3, 2) → 5
+(3, 3) → 6
+(3, 4) → 7
+(3, 5) → 8
+(3, 6) → 9
+(3, 7) → c(1, 0)
+(3, 8) → c(1, 1)
+(3, 9) → c(1, 2)
+(4, 0) → 4
+(4, 1) → 5
+(4, 2) → 6
+(4, 3) → 7
+(4, 4) → 8
+(4, 5) → 9
+(4, 6) → c(1, 0)
+(4, 7) → c(1, 1)
+(4, 8) → c(1, 2)
+(4, 9) → c(1, 3)
+(5, 0) → 5
+(5, 1) → 6
+(5, 2) → 7
+(5, 3) → 8
+(5, 4) → 9
+(5, 5) → c(1, 0)
+(5, 6) → c(1, 1)
+(5, 7) → c(1, 2)
+(5, 8) → c(1, 3)
+(5, 9) → c(1, 4)
+(6, 0) → 6
+(6, 1) → 7
+(6, 2) → 8
+(6, 3) → 9
+(6, 4) → c(1, 0)
+(6, 5) → c(1, 1)
+(6, 6) → c(1, 2)
+(6, 7) → c(1, 3)
+(6, 8) → c(1, 4)
+(6, 9) → c(1, 5)
+(7, 0) → 7
+(7, 1) → 8
+(7, 2) → 9
+(7, 3) → c(1, 0)
+(7, 4) → c(1, 1)
+(7, 5) → c(1, 2)
+(7, 6) → c(1, 3)
+(7, 7) → c(1, 4)
+(7, 8) → c(1, 5)
+(7, 9) → c(1, 6)
+(8, 0) → 8
+(8, 1) → 9
+(8, 2) → c(1, 0)
+(8, 3) → c(1, 1)
+(8, 4) → c(1, 2)
+(8, 5) → c(1, 3)
+(8, 6) → c(1, 4)
+(8, 7) → c(1, 5)
+(8, 8) → c(1, 6)
+(8, 9) → c(1, 7)
+(9, 0) → 9
+(9, 1) → c(1, 0)
+(9, 2) → c(1, 1)
+(9, 3) → c(1, 2)
+(9, 4) → c(1, 3)
+(9, 5) → c(1, 4)
+(9, 6) → c(1, 5)
+(9, 7) → c(1, 6)
+(9, 8) → c(1, 7)
+(9, 9) → c(1, 8)
+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))
c(0, x) → x
c(x, c(y, z)) → c(+(x, y), z)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

+(0, 0) → 0
+(0, 1) → 1
+(0, 2) → 2
+(0, 3) → 3
+(0, 4) → 4
+(0, 5) → 5
+(0, 6) → 6
+(0, 7) → 7
+(0, 8) → 8
+(0, 9) → 9
+(1, 0) → 1
+(1, 1) → 2
+(1, 2) → 3
+(1, 3) → 4
+(1, 4) → 5
+(1, 5) → 6
+(1, 6) → 7
+(1, 7) → 8
+(1, 8) → 9
+(1, 9) → c(1, 0)
+(2, 0) → 2
+(2, 1) → 3
+(2, 2) → 4
+(2, 3) → 5
+(2, 4) → 6
+(2, 5) → 7
+(2, 6) → 8
+(2, 7) → 9
+(2, 8) → c(1, 0)
+(2, 9) → c(1, 1)
+(3, 0) → 3
+(3, 1) → 4
+(3, 2) → 5
+(3, 3) → 6
+(3, 4) → 7
+(3, 5) → 8
+(3, 6) → 9
+(3, 7) → c(1, 0)
+(3, 8) → c(1, 1)
+(3, 9) → c(1, 2)
+(4, 0) → 4
+(4, 1) → 5
+(4, 2) → 6
+(4, 3) → 7
+(4, 4) → 8
+(4, 5) → 9
+(4, 6) → c(1, 0)
+(4, 7) → c(1, 1)
+(4, 8) → c(1, 2)
+(4, 9) → c(1, 3)
+(5, 0) → 5
+(5, 1) → 6
+(5, 2) → 7
+(5, 3) → 8
+(5, 4) → 9
+(5, 5) → c(1, 0)
+(5, 6) → c(1, 1)
+(5, 7) → c(1, 2)
+(5, 8) → c(1, 3)
+(5, 9) → c(1, 4)
+(6, 0) → 6
+(6, 1) → 7
+(6, 2) → 8
+(6, 3) → 9
+(6, 4) → c(1, 0)
+(6, 5) → c(1, 1)
+(6, 6) → c(1, 2)
+(6, 7) → c(1, 3)
+(6, 8) → c(1, 4)
+(6, 9) → c(1, 5)
+(7, 0) → 7
+(7, 1) → 8
+(7, 2) → 9
+(7, 3) → c(1, 0)
+(7, 4) → c(1, 1)
+(7, 5) → c(1, 2)
+(7, 6) → c(1, 3)
+(7, 7) → c(1, 4)
+(7, 8) → c(1, 5)
+(7, 9) → c(1, 6)
+(8, 0) → 8
+(8, 1) → 9
+(8, 2) → c(1, 0)
+(8, 3) → c(1, 1)
+(8, 4) → c(1, 2)
+(8, 5) → c(1, 3)
+(8, 6) → c(1, 4)
+(8, 7) → c(1, 5)
+(8, 8) → c(1, 6)
+(8, 9) → c(1, 7)
+(9, 0) → 9
+(9, 1) → c(1, 0)
+(9, 2) → c(1, 1)
+(9, 3) → c(1, 2)
+(9, 4) → c(1, 3)
+(9, 5) → c(1, 4)
+(9, 6) → c(1, 5)
+(9, 7) → c(1, 6)
+(9, 8) → c(1, 7)
+(9, 9) → c(1, 8)
+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))
c(0, x) → x
c(x, c(y, z)) → c(+(x, y), z)

Q is empty.

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

+(0, 0) → 0
+(0, 1) → 1
+(0, 2) → 2
+(0, 3) → 3
+(0, 4) → 4
+(0, 5) → 5
+(0, 6) → 6
+(0, 7) → 7
+(0, 8) → 8
+(0, 9) → 9
+(1, 0) → 1
+(1, 1) → 2
+(1, 2) → 3
+(1, 3) → 4
+(1, 4) → 5
+(1, 5) → 6
+(1, 6) → 7
+(1, 7) → 8
+(1, 8) → 9
+(1, 9) → c(1, 0)
+(2, 0) → 2
+(2, 1) → 3
+(2, 2) → 4
+(2, 3) → 5
+(2, 4) → 6
+(2, 5) → 7
+(2, 6) → 8
+(2, 7) → 9
+(2, 8) → c(1, 0)
+(2, 9) → c(1, 1)
+(3, 0) → 3
+(3, 1) → 4
+(3, 2) → 5
+(3, 3) → 6
+(3, 4) → 7
+(3, 5) → 8
+(3, 6) → 9
+(3, 7) → c(1, 0)
+(3, 8) → c(1, 1)
+(3, 9) → c(1, 2)
+(4, 0) → 4
+(4, 1) → 5
+(4, 2) → 6
+(4, 3) → 7
+(4, 4) → 8
+(4, 5) → 9
+(4, 6) → c(1, 0)
+(4, 7) → c(1, 1)
+(4, 8) → c(1, 2)
+(4, 9) → c(1, 3)
+(5, 0) → 5
+(5, 1) → 6
+(5, 2) → 7
+(5, 3) → 8
+(5, 4) → 9
+(5, 5) → c(1, 0)
+(5, 6) → c(1, 1)
+(5, 7) → c(1, 2)
+(5, 8) → c(1, 3)
+(5, 9) → c(1, 4)
+(6, 0) → 6
+(6, 1) → 7
+(6, 2) → 8
+(6, 3) → 9
+(6, 4) → c(1, 0)
+(6, 5) → c(1, 1)
+(6, 6) → c(1, 2)
+(6, 7) → c(1, 3)
+(6, 8) → c(1, 4)
+(6, 9) → c(1, 5)
+(7, 0) → 7
+(7, 1) → 8
+(7, 2) → 9
+(7, 3) → c(1, 0)
+(7, 4) → c(1, 1)
+(7, 5) → c(1, 2)
+(7, 6) → c(1, 3)
+(7, 7) → c(1, 4)
+(7, 8) → c(1, 5)
+(7, 9) → c(1, 6)
+(8, 0) → 8
+(8, 1) → 9
+(8, 2) → c(1, 0)
+(8, 3) → c(1, 1)
+(8, 4) → c(1, 2)
+(8, 5) → c(1, 3)
+(8, 6) → c(1, 4)
+(8, 7) → c(1, 5)
+(8, 8) → c(1, 6)
+(8, 9) → c(1, 7)
+(9, 0) → 9
+(9, 1) → c(1, 0)
+(9, 2) → c(1, 1)
+(9, 3) → c(1, 2)
+(9, 4) → c(1, 3)
+(9, 5) → c(1, 4)
+(9, 6) → c(1, 5)
+(9, 7) → c(1, 6)
+(9, 8) → c(1, 7)
+(9, 9) → c(1, 8)
+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))
c(0, x) → x
c(x, c(y, z)) → c(+(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:

+(0, 0) → 0
+(0, 1) → 1
+(0, 2) → 2
+(0, 3) → 3
+(0, 4) → 4
+(0, 5) → 5
+(0, 6) → 6
+(0, 7) → 7
+(0, 8) → 8
+(0, 9) → 9
+(1, 0) → 1
+(1, 1) → 2
+(1, 2) → 3
+(1, 3) → 4
+(1, 5) → 6
+(1, 6) → 7
+(1, 7) → 8
+(1, 8) → 9
+(2, 0) → 2
+(2, 1) → 3
+(2, 2) → 4
+(2, 5) → 7
+(2, 6) → 8
+(2, 7) → 9
+(3, 0) → 3
+(3, 1) → 4
+(3, 5) → 8
+(3, 6) → 9
+(4, 0) → 4
+(4, 5) → 9
+(5, 0) → 5
+(5, 1) → 6
+(5, 2) → 7
+(5, 3) → 8
+(5, 4) → 9
+(5, 5) → c(1, 0)
+(5, 6) → c(1, 1)
+(5, 7) → c(1, 2)
+(5, 8) → c(1, 3)
+(5, 9) → c(1, 4)
+(6, 0) → 6
+(6, 1) → 7
+(6, 2) → 8
+(6, 3) → 9
+(6, 5) → c(1, 1)
+(6, 6) → c(1, 2)
+(6, 7) → c(1, 3)
+(6, 8) → c(1, 4)
+(7, 0) → 7
+(7, 1) → 8
+(7, 2) → 9
+(7, 5) → c(1, 2)
+(7, 6) → c(1, 3)
+(7, 7) → c(1, 4)
+(8, 0) → 8
+(8, 1) → 9
+(8, 5) → c(1, 3)
+(8, 6) → c(1, 4)
+(9, 0) → 9
+(9, 5) → c(1, 4)
c(0, x) → x
c(x, c(y, z)) → c(+(x, y), z)

Used ordering:
Polynomial interpretation [25]:

POL(+(x₁, x₂)) = x₁ + x₂
POL(0) = 1
POL(1) = 1
POL(2) = 1
POL(3) = 1
POL(4) = 1
POL(5) = 2
POL(6) = 2
POL(7) = 2
POL(8) = 2
POL(9) = 2
POL(c(x₁, x₂)) = 1 + x₁ + x₂

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

+(1, 4) → 5
+(1, 9) → c(1, 0)
+(2, 3) → 5
+(2, 4) → 6
+(2, 8) → c(1, 0)
+(2, 9) → c(1, 1)
+(3, 2) → 5
+(3, 3) → 6
+(3, 4) → 7
+(3, 7) → c(1, 0)
+(3, 8) → c(1, 1)
+(3, 9) → c(1, 2)
+(4, 1) → 5
+(4, 2) → 6
+(4, 3) → 7
+(4, 4) → 8
+(4, 6) → c(1, 0)
+(4, 7) → c(1, 1)
+(4, 8) → c(1, 2)
+(4, 9) → c(1, 3)
+(6, 4) → c(1, 0)
+(6, 9) → c(1, 5)
+(7, 3) → c(1, 0)
+(7, 4) → c(1, 1)
+(7, 8) → c(1, 5)
+(7, 9) → c(1, 6)
+(8, 2) → c(1, 0)
+(8, 3) → c(1, 1)
+(8, 4) → c(1, 2)
+(8, 7) → c(1, 5)
+(8, 8) → c(1, 6)
+(8, 9) → c(1, 7)
+(9, 1) → c(1, 0)
+(9, 2) → c(1, 1)
+(9, 3) → c(1, 2)
+(9, 4) → c(1, 3)
+(9, 6) → c(1, 5)
+(9, 7) → c(1, 6)
+(9, 8) → c(1, 7)
+(9, 9) → c(1, 8)
+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))

Q is empty.

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

+(1, 4) → 5
+(1, 9) → c(1, 0)
+(2, 3) → 5
+(2, 4) → 6
+(2, 8) → c(1, 0)
+(2, 9) → c(1, 1)
+(3, 2) → 5
+(3, 3) → 6
+(3, 4) → 7
+(3, 7) → c(1, 0)
+(3, 8) → c(1, 1)
+(3, 9) → c(1, 2)
+(4, 1) → 5
+(4, 2) → 6
+(4, 3) → 7
+(4, 4) → 8
+(4, 6) → c(1, 0)
+(4, 7) → c(1, 1)
+(4, 8) → c(1, 2)
+(4, 9) → c(1, 3)
+(6, 4) → c(1, 0)
+(6, 9) → c(1, 5)
+(7, 3) → c(1, 0)
+(7, 4) → c(1, 1)
+(7, 8) → c(1, 5)
+(7, 9) → c(1, 6)
+(8, 2) → c(1, 0)
+(8, 3) → c(1, 1)
+(8, 4) → c(1, 2)
+(8, 7) → c(1, 5)
+(8, 8) → c(1, 6)
+(8, 9) → c(1, 7)
+(9, 1) → c(1, 0)
+(9, 2) → c(1, 1)
+(9, 3) → c(1, 2)
+(9, 4) → c(1, 3)
+(9, 6) → c(1, 5)
+(9, 7) → c(1, 6)
+(9, 8) → c(1, 7)
+(9, 9) → c(1, 8)
+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(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:

+(1, 4) → 5
+(1, 9) → c(1, 0)
+(2, 3) → 5
+(2, 4) → 6
+(2, 8) → c(1, 0)
+(2, 9) → c(1, 1)
+(3, 2) → 5
+(3, 3) → 6
+(3, 4) → 7
+(3, 7) → c(1, 0)
+(3, 8) → c(1, 1)
+(3, 9) → c(1, 2)
+(4, 1) → 5
+(4, 2) → 6
+(4, 3) → 7
+(4, 4) → 8
+(4, 6) → c(1, 0)
+(4, 7) → c(1, 1)
+(4, 8) → c(1, 2)
+(4, 9) → c(1, 3)
+(6, 4) → c(1, 0)
+(6, 9) → c(1, 5)
+(7, 3) → c(1, 0)
+(7, 4) → c(1, 1)
+(7, 8) → c(1, 5)
+(7, 9) → c(1, 6)
+(8, 2) → c(1, 0)
+(8, 3) → c(1, 1)
+(8, 4) → c(1, 2)
+(8, 7) → c(1, 5)
+(8, 8) → c(1, 6)
+(8, 9) → c(1, 7)
+(9, 1) → c(1, 0)
+(9, 2) → c(1, 1)
+(9, 3) → c(1, 2)
+(9, 4) → c(1, 3)
+(9, 6) → c(1, 5)
+(9, 7) → c(1, 6)
+(9, 8) → c(1, 7)
+(9, 9) → c(1, 8)
+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))

Used ordering:
Polynomial interpretation [25]:

POL(+(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(0) = 2
POL(1) = 0
POL(2) = 2
POL(3) = 2
POL(4) = 2
POL(5) = 2
POL(6) = 2
POL(7) = 2
POL(8) = 1
POL(9) = 1
POL(c(x₁, x₂)) = 1 + 2·x₁ + x₂

↳ 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.