Termination w.r.t. Q proof of /tmp/tmpT7F6yn/t005.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 105 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

C(x, c(y, z)) → +¹(x, y)
+¹(x, c(y, z)) → C(y, +(x, z))
+¹(c(x, y), z) → +¹(y, z)
+¹(x, c(y, z)) → +¹(x, z)
C(x, c(y, z)) → C(+(x, y), z)
+¹(c(x, y), z) → C(x, +(y, z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

          ↳ QDP

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

C(x, c(y, z)) → +¹(x, y)
+¹(x, c(y, z)) → C(y, +(x, z))
+¹(x, c(y, z)) → +¹(x, z)
+¹(c(x, y), z) → +¹(y, z)
C(x, c(y, z)) → C(+(x, y), z)
+¹(c(x, y), z) → C(x, +(y, z))

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.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

          ↳ QDP

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

C(x, c(y, z)) → +¹(x, y)
+¹(x, c(y, z)) → C(y, +(x, z))
+¹(c(x, y), z) → +¹(y, z)
+¹(x, c(y, z)) → +¹(x, z)
C(x, c(y, z)) → C(+(x, y), z)
+¹(c(x, y), z) → C(x, +(y, z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ RuleRemovalProof

          ↳ QDP

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

C(x, c(y, z)) → +¹(x, y)
+¹(x, c(y, z)) → C(y, +(x, z))
+¹(x, c(y, z)) → +¹(x, z)
+¹(c(x, y), z) → +¹(y, z)
C(x, c(y, z)) → C(+(x, y), z)
+¹(c(x, y), z) → C(x, +(y, z))

The TRS R consists of the following rules:

+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))
c(x, c(y, z)) → c(+(x, y), z)

C(x, c(y, z)) → +¹(x, y)
+¹(x, c(y, z)) → C(y, +(x, z))
+¹(x, c(y, z)) → +¹(x, z)
+¹(c(x, y), z) → +¹(y, z)
C(x, c(y, z)) → C(+(x, y), z)
+¹(c(x, y), z) → C(x, +(y, z))

Strictly oriented rules of the TRS R:

c(x, c(y, z)) → c(+(x, y), z)

Used ordering: POLO with Polynomial interpretation [25]:

POL(+(x₁, x₂)) = x₁ + x₂
POL(+¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(C(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(c(x₁, x₂)) = 2 + 2·x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, c(y, z)) → c(y, +(x, z))
+(c(x, y), z) → c(x, +(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

*¹(c(x, y), z) → *¹(x, z)
*¹(x, c(y, z)) → *¹(x, z)
*¹(c(x, y), z) → *¹(y, z)
*¹(x, c(y, z)) → *¹(x, y)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

*¹(x, c(y, z)) → *¹(x, z)
*¹(c(x, y), z) → *¹(x, z)
*¹(c(x, y), z) → *¹(y, z)
*¹(x, c(y, z)) → *¹(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

*¹(c(x, y), z) → *¹(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(x, c(y, z)) → *¹(x, z)
The graph contains the following edges 1 >= 1, 2 > 2
*¹(c(x, y), z) → *¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(x, c(y, z)) → *¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2