YES

The TRS could be proven terminating. The proof took 4052 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (1037ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (2216ms).
 |    | – Problem 3 was processed with processor DependencyGraph (13ms).
 |    |    | – Problem 4 was processed with processor PolynomialLinearRange4iUR (554ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite 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)xc(x, c(y, z))c(+(x, y), z)

Original Signature

Termination of terms over the following signature is verified: 3, 2, 1, 0, 7, 6, c, 5, 4, +, 9, 8

Strategy


The following SCCs where found

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

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite 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)xc(x, c(y, z))c(+(x, y), z)

Original Signature

Termination of terms over the following signature is verified: 3, 2, 1, 0, 7, 6, c, 5, 4, +, 9, 8

Strategy


Polynomial Interpretation

Improved Usable rules

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

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

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

Problem 3: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite 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)xc(x, c(y, z))c(+(x, y), z)

Original Signature

Termination of terms over the following signature is verified: 3, 2, 1, 0, 7, c, 6, 5, 4, +, 9, 8

Strategy


The following SCCs where found

c#(x, c(y, z)) → c#(+(x, y), z)

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite 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)xc(x, c(y, z))c(+(x, y), z)

Original Signature

Termination of terms over the following signature is verified: 3, 2, 1, 0, 7, c, 6, 5, 4, +, 9, 8

Strategy


Polynomial Interpretation

Improved Usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

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