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)