(0) Obligation:
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.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(1, 9) → C(1, 0)
+1(2, 8) → C(1, 0)
+1(2, 9) → C(1, 1)
+1(3, 7) → C(1, 0)
+1(3, 8) → C(1, 1)
+1(3, 9) → C(1, 2)
+1(4, 6) → C(1, 0)
+1(4, 7) → C(1, 1)
+1(4, 8) → C(1, 2)
+1(4, 9) → C(1, 3)
+1(5, 5) → C(1, 0)
+1(5, 6) → C(1, 1)
+1(5, 7) → C(1, 2)
+1(5, 8) → C(1, 3)
+1(5, 9) → C(1, 4)
+1(6, 4) → C(1, 0)
+1(6, 5) → C(1, 1)
+1(6, 6) → C(1, 2)
+1(6, 7) → C(1, 3)
+1(6, 8) → C(1, 4)
+1(6, 9) → C(1, 5)
+1(7, 3) → C(1, 0)
+1(7, 4) → C(1, 1)
+1(7, 5) → C(1, 2)
+1(7, 6) → C(1, 3)
+1(7, 7) → C(1, 4)
+1(7, 8) → C(1, 5)
+1(7, 9) → C(1, 6)
+1(8, 2) → C(1, 0)
+1(8, 3) → C(1, 1)
+1(8, 4) → C(1, 2)
+1(8, 5) → C(1, 3)
+1(8, 6) → C(1, 4)
+1(8, 7) → C(1, 5)
+1(8, 8) → C(1, 6)
+1(8, 9) → C(1, 7)
+1(9, 1) → C(1, 0)
+1(9, 2) → C(1, 1)
+1(9, 3) → C(1, 2)
+1(9, 4) → C(1, 3)
+1(9, 5) → C(1, 4)
+1(9, 6) → C(1, 5)
+1(9, 7) → C(1, 6)
+1(9, 8) → C(1, 7)
+1(9, 9) → C(1, 8)
+1(x, c(y, z)) → C(y, +(x, z))
+1(x, c(y, z)) → +1(x, z)
+1(c(x, y), z) → C(x, +(y, z))
+1(c(x, y), z) → +1(y, z)
C(x, c(y, z)) → C(+(x, y), z)
C(x, c(y, z)) → +1(x, y)
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.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 45 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(x, c(y, z)) → C(y, +(x, z))
C(x, c(y, z)) → C(+(x, y), z)
C(x, c(y, z)) → +1(x, y)
+1(x, c(y, z)) → +1(x, z)
+1(c(x, y), z) → C(x, +(y, z))
+1(c(x, y), z) → +1(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.