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:

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, y), z)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

+12(7, 3) -> C2(1, 0)
+12(6, 4) -> C2(1, 0)
+12(1, 9) -> C2(1, 0)
+12(2, 8) -> C2(1, 0)
+12(3, 7) -> C2(1, 0)
+12(4, 6) -> C2(1, 0)
+12(5, 5) -> C2(1, 0)
+12(8, 2) -> C2(1, 0)
+12(9, 1) -> C2(1, 0)
+12(x, c2(y, z)) -> +12(x, z)
+12(7, 5) -> C2(1, 2)
+12(6, 6) -> C2(1, 2)
+12(3, 9) -> C2(1, 2)
+12(4, 8) -> C2(1, 2)
+12(5, 7) -> C2(1, 2)
+12(8, 4) -> C2(1, 2)
+12(9, 3) -> C2(1, 2)
+12(7, 6) -> C2(1, 3)
+12(6, 7) -> C2(1, 3)
+12(5, 8) -> C2(1, 3)
+12(4, 9) -> C2(1, 3)
+12(8, 5) -> C2(1, 3)
+12(9, 4) -> C2(1, 3)
+12(x, c2(y, z)) -> C2(y, +2(x, z))
+12(c2(x, y), z) -> C2(x, +2(y, z))
C2(x, c2(y, z)) -> +12(x, y)
+12(7, 9) -> C2(1, 6)
+12(8, 8) -> C2(1, 6)
+12(9, 7) -> C2(1, 6)
+12(c2(x, y), z) -> +12(y, z)
C2(x, c2(y, z)) -> C2(+2(x, y), z)
+12(6, 8) -> C2(1, 4)
+12(5, 9) -> C2(1, 4)
+12(7, 7) -> C2(1, 4)
+12(8, 6) -> C2(1, 4)
+12(9, 5) -> C2(1, 4)
+12(9, 9) -> C2(1, 8)
+12(6, 9) -> C2(1, 5)
+12(7, 8) -> C2(1, 5)
+12(8, 7) -> C2(1, 5)
+12(9, 6) -> C2(1, 5)
+12(7, 4) -> C2(1, 1)
+12(6, 5) -> C2(1, 1)
+12(2, 9) -> C2(1, 1)
+12(3, 8) -> C2(1, 1)
+12(4, 7) -> C2(1, 1)
+12(5, 6) -> C2(1, 1)
+12(8, 3) -> C2(1, 1)
+12(9, 2) -> C2(1, 1)
+12(8, 9) -> C2(1, 7)
+12(9, 8) -> C2(1, 7)

The TRS R consists of the following rules:

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, 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:

+12(7, 3) -> C2(1, 0)
+12(6, 4) -> C2(1, 0)
+12(1, 9) -> C2(1, 0)
+12(2, 8) -> C2(1, 0)
+12(3, 7) -> C2(1, 0)
+12(4, 6) -> C2(1, 0)
+12(5, 5) -> C2(1, 0)
+12(8, 2) -> C2(1, 0)
+12(9, 1) -> C2(1, 0)
+12(x, c2(y, z)) -> +12(x, z)
+12(7, 5) -> C2(1, 2)
+12(6, 6) -> C2(1, 2)
+12(3, 9) -> C2(1, 2)
+12(4, 8) -> C2(1, 2)
+12(5, 7) -> C2(1, 2)
+12(8, 4) -> C2(1, 2)
+12(9, 3) -> C2(1, 2)
+12(7, 6) -> C2(1, 3)
+12(6, 7) -> C2(1, 3)
+12(5, 8) -> C2(1, 3)
+12(4, 9) -> C2(1, 3)
+12(8, 5) -> C2(1, 3)
+12(9, 4) -> C2(1, 3)
+12(x, c2(y, z)) -> C2(y, +2(x, z))
+12(c2(x, y), z) -> C2(x, +2(y, z))
C2(x, c2(y, z)) -> +12(x, y)
+12(7, 9) -> C2(1, 6)
+12(8, 8) -> C2(1, 6)
+12(9, 7) -> C2(1, 6)
+12(c2(x, y), z) -> +12(y, z)
C2(x, c2(y, z)) -> C2(+2(x, y), z)
+12(6, 8) -> C2(1, 4)
+12(5, 9) -> C2(1, 4)
+12(7, 7) -> C2(1, 4)
+12(8, 6) -> C2(1, 4)
+12(9, 5) -> C2(1, 4)
+12(9, 9) -> C2(1, 8)
+12(6, 9) -> C2(1, 5)
+12(7, 8) -> C2(1, 5)
+12(8, 7) -> C2(1, 5)
+12(9, 6) -> C2(1, 5)
+12(7, 4) -> C2(1, 1)
+12(6, 5) -> C2(1, 1)
+12(2, 9) -> C2(1, 1)
+12(3, 8) -> C2(1, 1)
+12(4, 7) -> C2(1, 1)
+12(5, 6) -> C2(1, 1)
+12(8, 3) -> C2(1, 1)
+12(9, 2) -> C2(1, 1)
+12(8, 9) -> C2(1, 7)
+12(9, 8) -> C2(1, 7)

The TRS R consists of the following rules:

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 45 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

+12(c2(x, y), z) -> +12(y, z)
+12(x, c2(y, z)) -> +12(x, z)
C2(x, c2(y, z)) -> +12(x, y)
C2(x, c2(y, z)) -> C2(+2(x, y), z)
+12(x, c2(y, z)) -> C2(y, +2(x, z))
+12(c2(x, y), z) -> C2(x, +2(y, z))

The TRS R consists of the following rules:

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+12(c2(x, y), z) -> +12(y, z)
+12(x, c2(y, z)) -> +12(x, z)
+12(x, c2(y, z)) -> C2(y, +2(x, z))
+12(c2(x, y), z) -> C2(x, +2(y, z))
The remaining pairs can at least be oriented weakly.

C2(x, c2(y, z)) -> +12(x, y)
C2(x, c2(y, z)) -> C2(+2(x, y), z)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = max{0, x1 + x2 - 2}


POL( c2(x1, x2) ) = x1 + x2 + 3


POL( C2(x1, x2) ) = max{0, x1 + x2 - 3}


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


POL( 1 ) = 0


POL( 6 ) = 1


POL( 7 ) = 2


POL( 5 ) = 1


POL( 4 ) = 1


POL( 8 ) = 2


POL( 3 ) = 0


POL( 9 ) = 2


POL( 0 ) = 0


POL( 2 ) = 0



The following usable rules [14] were oriented:

+2(2, 9) -> c2(1, 1)
+2(5, 3) -> 8
+2(6, 1) -> 7
+2(0, 5) -> 5
+2(2, 4) -> 6
+2(6, 0) -> 6
+2(0, 6) -> 6
+2(7, 0) -> 7
+2(9, 5) -> c2(1, 4)
+2(3, 1) -> 4
+2(0, 3) -> 3
+2(5, 2) -> 7
+2(6, 7) -> c2(1, 3)
+2(5, 7) -> c2(1, 2)
+2(0, 8) -> 8
+2(6, 4) -> c2(1, 0)
+2(0, 0) -> 0
+2(3, 3) -> 6
+2(2, 7) -> 9
+2(7, 1) -> 8
+2(6, 6) -> c2(1, 2)
+2(2, 6) -> 8
+2(0, 9) -> 9
+2(9, 9) -> c2(1, 8)
+2(0, 4) -> 4
+2(1, 0) -> 1
+2(7, 4) -> c2(1, 1)
+2(6, 9) -> c2(1, 5)
+2(1, 3) -> 4
+2(8, 8) -> c2(1, 6)
+2(1, 2) -> 3
+2(3, 9) -> c2(1, 2)
+2(6, 2) -> 8
c2(x, c2(y, z)) -> c2(+2(x, y), z)
+2(3, 0) -> 3
+2(7, 5) -> c2(1, 2)
+2(1, 1) -> 2
+2(8, 9) -> c2(1, 7)
+2(8, 1) -> 9
+2(2, 2) -> 4
+2(4, 9) -> c2(1, 3)
+2(0, 2) -> 2
+2(2, 8) -> c2(1, 0)
+2(0, 1) -> 1
+2(5, 0) -> 5
+2(3, 8) -> c2(1, 1)
+2(6, 3) -> 9
+2(9, 1) -> c2(1, 0)
+2(5, 1) -> 6
+2(9, 0) -> 9
+2(8, 0) -> 8
+2(4, 8) -> c2(1, 2)
+2(4, 6) -> c2(1, 0)
+2(4, 0) -> 4
+2(9, 6) -> c2(1, 5)
+2(c2(x, y), z) -> c2(x, +2(y, z))
+2(7, 7) -> c2(1, 4)
+2(1, 8) -> 9
+2(2, 3) -> 5
+2(8, 7) -> c2(1, 5)
+2(8, 2) -> c2(1, 0)
c2(0, x) -> x
+2(7, 6) -> c2(1, 3)
+2(2, 0) -> 2
+2(8, 3) -> c2(1, 1)
+2(4, 7) -> c2(1, 1)
+2(4, 1) -> 5
+2(2, 1) -> 3
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(4, 2) -> 6
+2(3, 6) -> 9
+2(1, 4) -> 5
+2(7, 2) -> 9
+2(8, 6) -> c2(1, 4)
+2(1, 9) -> c2(1, 0)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(1, 5) -> 6
+2(7, 3) -> c2(1, 0)
+2(4, 3) -> 7
+2(6, 8) -> c2(1, 4)
+2(3, 2) -> 5
+2(5, 4) -> 9
+2(9, 3) -> c2(1, 2)
+2(3, 7) -> c2(1, 0)
+2(4, 5) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 9) -> c2(1, 4)
+2(4, 4) -> 8
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(9, 4) -> c2(1, 3)
+2(7, 8) -> c2(1, 5)
+2(2, 5) -> 7
+2(1, 7) -> 8
+2(5, 6) -> c2(1, 1)
+2(6, 5) -> c2(1, 1)
+2(9, 2) -> c2(1, 1)
+2(3, 4) -> 7
+2(0, 7) -> 7
+2(7, 9) -> c2(1, 6)
+2(5, 8) -> c2(1, 3)
+2(3, 5) -> 8
+2(1, 6) -> 7



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

C2(x, c2(y, z)) -> +12(x, y)
C2(x, c2(y, z)) -> C2(+2(x, y), z)

The TRS R consists of the following rules:

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

C2(x, c2(y, z)) -> C2(+2(x, y), z)

The TRS R consists of the following rules:

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


C2(x, c2(y, z)) -> C2(+2(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( C2(x1, x2) ) = max{0, x2 - 2}


POL( c2(x1, x2) ) = x2 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

+2(0, 0) -> 0
+2(0, 1) -> 1
+2(0, 2) -> 2
+2(0, 3) -> 3
+2(0, 4) -> 4
+2(0, 5) -> 5
+2(0, 6) -> 6
+2(0, 7) -> 7
+2(0, 8) -> 8
+2(0, 9) -> 9
+2(1, 0) -> 1
+2(1, 1) -> 2
+2(1, 2) -> 3
+2(1, 3) -> 4
+2(1, 4) -> 5
+2(1, 5) -> 6
+2(1, 6) -> 7
+2(1, 7) -> 8
+2(1, 8) -> 9
+2(1, 9) -> c2(1, 0)
+2(2, 0) -> 2
+2(2, 1) -> 3
+2(2, 2) -> 4
+2(2, 3) -> 5
+2(2, 4) -> 6
+2(2, 5) -> 7
+2(2, 6) -> 8
+2(2, 7) -> 9
+2(2, 8) -> c2(1, 0)
+2(2, 9) -> c2(1, 1)
+2(3, 0) -> 3
+2(3, 1) -> 4
+2(3, 2) -> 5
+2(3, 3) -> 6
+2(3, 4) -> 7
+2(3, 5) -> 8
+2(3, 6) -> 9
+2(3, 7) -> c2(1, 0)
+2(3, 8) -> c2(1, 1)
+2(3, 9) -> c2(1, 2)
+2(4, 0) -> 4
+2(4, 1) -> 5
+2(4, 2) -> 6
+2(4, 3) -> 7
+2(4, 4) -> 8
+2(4, 5) -> 9
+2(4, 6) -> c2(1, 0)
+2(4, 7) -> c2(1, 1)
+2(4, 8) -> c2(1, 2)
+2(4, 9) -> c2(1, 3)
+2(5, 0) -> 5
+2(5, 1) -> 6
+2(5, 2) -> 7
+2(5, 3) -> 8
+2(5, 4) -> 9
+2(5, 5) -> c2(1, 0)
+2(5, 6) -> c2(1, 1)
+2(5, 7) -> c2(1, 2)
+2(5, 8) -> c2(1, 3)
+2(5, 9) -> c2(1, 4)
+2(6, 0) -> 6
+2(6, 1) -> 7
+2(6, 2) -> 8
+2(6, 3) -> 9
+2(6, 4) -> c2(1, 0)
+2(6, 5) -> c2(1, 1)
+2(6, 6) -> c2(1, 2)
+2(6, 7) -> c2(1, 3)
+2(6, 8) -> c2(1, 4)
+2(6, 9) -> c2(1, 5)
+2(7, 0) -> 7
+2(7, 1) -> 8
+2(7, 2) -> 9
+2(7, 3) -> c2(1, 0)
+2(7, 4) -> c2(1, 1)
+2(7, 5) -> c2(1, 2)
+2(7, 6) -> c2(1, 3)
+2(7, 7) -> c2(1, 4)
+2(7, 8) -> c2(1, 5)
+2(7, 9) -> c2(1, 6)
+2(8, 0) -> 8
+2(8, 1) -> 9
+2(8, 2) -> c2(1, 0)
+2(8, 3) -> c2(1, 1)
+2(8, 4) -> c2(1, 2)
+2(8, 5) -> c2(1, 3)
+2(8, 6) -> c2(1, 4)
+2(8, 7) -> c2(1, 5)
+2(8, 8) -> c2(1, 6)
+2(8, 9) -> c2(1, 7)
+2(9, 0) -> 9
+2(9, 1) -> c2(1, 0)
+2(9, 2) -> c2(1, 1)
+2(9, 3) -> c2(1, 2)
+2(9, 4) -> c2(1, 3)
+2(9, 5) -> c2(1, 4)
+2(9, 6) -> c2(1, 5)
+2(9, 7) -> c2(1, 6)
+2(9, 8) -> c2(1, 7)
+2(9, 9) -> c2(1, 8)
+2(x, c2(y, z)) -> c2(y, +2(x, z))
+2(c2(x, y), z) -> c2(x, +2(y, z))
c2(0, x) -> x
c2(x, c2(y, z)) -> c2(+2(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.