The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(0) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(U22(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U22(dout(z0), z1, z2, z3) → c4
U31(dout(z0), z1, z2) → c5(U32(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U32(dout(z0), z1, z2, z3) → c6
U41(dout(z0), z1) → c7(U42(din(der(z0)), z1, z0), DIN(der(z0)))
U42(dout(z0), z1, z2) → c8

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(U22(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U22(dout(z0), z1, z2, z3) → c4
U31(dout(z0), z1, z2) → c5(U32(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U32(dout(z0), z1, z2, z3) → c6
U41(dout(z0), z1) → c7(U42(din(der(z0)), z1, z0), DIN(der(z0)))
U42(dout(z0), z1, z2) → c8

K tuples:none
Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U22, U31, U32, U41, U42

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing nodes:

U32(dout(z0), z1, z2, z3) → c6
U42(dout(z0), z1, z2) → c8
U22(dout(z0), z1, z2, z3) → c4

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(U22(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(U32(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U41(dout(z0), z1) → c7(U42(din(der(z0)), z1, z0), DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(U22(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(U32(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U41(dout(z0), z1) → c7(U42(din(der(z0)), z1, z0), DIN(der(z0)))

K tuples:none
Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

K tuples:none
Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))

We considered the (Usable) Rules:

u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
u42(dout(z0), z1, z2) → dout(z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
din(der(der(z0))) → u41(din(der(z0)), z0)

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = 0   
POL(U21(x₁, x₂, x₃)) = [2]x₁   
POL(U31(x₁, x₂, x₃)) = 0   
POL(U41(x₁, x₂)) = 0   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = 0   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = [3] + x₁   
POL(times(x₁, x₂)) = [2]   
POL(u21(x₁, x₂, x₃)) = [3]x₁   
POL(u22(x₁, x₂, x₃, x₄)) = [2] + [3]x₁ + [2]x₄   
POL(u31(x₁, x₂, x₃)) = [2]x₁   
POL(u32(x₁, x₂, x₃, x₄)) = [2] + [3]x₁ + x₄   
POL(u41(x₁, x₂)) = 0   
POL(u42(x₁, x₂, x₃)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

K tuples:

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))

We considered the (Usable) Rules:

u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
u42(dout(z0), z1, z2) → dout(z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
din(der(der(z0))) → u41(din(der(z0)), z0)

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = 0   
POL(U21(x₁, x₂, x₃)) = x₁   
POL(U31(x₁, x₂, x₃)) = [2]x₁   
POL(U41(x₁, x₂)) = 0   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = 0   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = [3] + x₁   
POL(times(x₁, x₂)) = [1]   
POL(u21(x₁, x₂, x₃)) = [3]x₁   
POL(u22(x₁, x₂, x₃, x₄)) = [2] + [3]x₁ + [2]x₄   
POL(u31(x₁, x₂, x₃)) = 0   
POL(u32(x₁, x₂, x₃, x₄)) = [3]x₁   
POL(u41(x₁, x₂)) = 0   
POL(u42(x₁, x₂, x₃)) = [2]x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

K tuples:

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

U41(dout(z0), z1) → c7(DIN(der(z0)))

We considered the (Usable) Rules:

u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
u42(dout(z0), z1, z2) → dout(z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
din(der(der(z0))) → u41(din(der(z0)), z0)

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = 0   
POL(U21(x₁, x₂, x₃)) = 0   
POL(U31(x₁, x₂, x₃)) = x₁   
POL(U41(x₁, x₂)) = x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = 0   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [1] + x₁   
POL(plus(x₁, x₂)) = 0   
POL(times(x₁, x₂)) = 0   
POL(u21(x₁, x₂, x₃)) = x₁   
POL(u22(x₁, x₂, x₃, x₄)) = [1] + x₁   
POL(u31(x₁, x₂, x₃)) = x₁   
POL(u32(x₁, x₂, x₃, x₄)) = [1] + x₁ + x₄   
POL(u41(x₁, x₂)) = 0   
POL(u42(x₁, x₂, x₃)) = x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

K tuples:

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))

We considered the (Usable) Rules:

u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
u42(dout(z0), z1, z2) → dout(z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
din(der(der(z0))) → u41(din(der(z0)), z0)

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = x₁   
POL(U21(x₁, x₂, x₃)) = x₁·x₃   
POL(U31(x₁, x₂, x₃)) = x₃   
POL(U41(x₁, x₂)) = x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = x₁   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [1] + x₁   
POL(plus(x₁, x₂)) = [2] + x₁   
POL(times(x₁, x₂)) = x₁ + x₂   
POL(u21(x₁, x₂, x₃)) = [2]x₁·x₃ + [2]x₁·x₂   
POL(u22(x₁, x₂, x₃, x₄)) = [2]x₁ + x₂ + [2]x₃ + x₃·x₄ + x₂·x₄ + x₁·x₄ + x₁² + [2]x₁·x₃   
POL(u31(x₁, x₂, x₃)) = [2]x₁ + [2]x₁·x₃ + x₁·x₂   
POL(u32(x₁, x₂, x₃, x₄)) = [2] + x₁ + x₂ + x₃ + [2]x₃·x₄ + x₂·x₄   
POL(u41(x₁, x₂)) = x₁²   
POL(u42(x₁, x₂, x₃)) = x₃ + x₁·x₃ + x₁²   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

K tuples:

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))

We considered the (Usable) Rules:

u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
u42(dout(z0), z1, z2) → dout(z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
din(der(der(z0))) → u41(din(der(z0)), z0)

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = x₁²   
POL(U21(x₁, x₂, x₃)) = x₃² + [2]x₂·x₃ + x₁·x₃   
POL(U31(x₁, x₂, x₃)) = x₂ + x₃² + [2]x₂·x₃   
POL(U41(x₁, x₂)) = [2]x₁²   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = x₁   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(times(x₁, x₂)) = [1] + x₁ + x₂   
POL(u21(x₁, x₂, x₃)) = x₁ + [2]x₁·x₃ + [2]x₁·x₂   
POL(u22(x₁, x₂, x₃, x₄)) = [1] + [2]x₁ + x₃·x₄ + x₁·x₄ + [2]x₁·x₂ + [2]x₁·x₃   
POL(u31(x₁, x₂, x₃)) = [2]x₁·x₃ + [2]x₁²   
POL(u32(x₁, x₂, x₃, x₄)) = [2] + [2]x₁ + x₄² + x₃·x₄ + [2]x₁·x₄ + x₁² + [2]x₁·x₂ + [2]x₁·x₃   
POL(u41(x₁, x₂)) = [2]x₁·x₂   
POL(u42(x₁, x₂, x₃)) = [2]x₁ + x₂ + x₂·x₃ + x₁·x₃ + x₁·x₂   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

K tuples:

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

We considered the (Usable) Rules:

u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
u42(dout(z0), z1, z2) → dout(z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
din(der(der(z0))) → u41(din(der(z0)), z0)

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = x₁   
POL(U21(x₁, x₂, x₃)) = x₁·x₃ + [2]x₁² + x₁·x₂   
POL(U31(x₁, x₂, x₃)) = x₁ + x₁·x₃   
POL(U41(x₁, x₂)) = [2]x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = [2] + x₁   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = x₁   
POL(times(x₁, x₂)) = x₁ + x₂   
POL(u21(x₁, x₂, x₃)) = [2]x₁·x₃ + x₁² + [2]x₁·x₂   
POL(u22(x₁, x₂, x₃, x₄)) = [2] + x₁ + x₂ + x₃ + [2]x₄ + x₄² + x₃·x₄ + x₂·x₄ + [2]x₁·x₄ + x₁²   
POL(u31(x₁, x₂, x₃)) = x₁ + [2]x₁·x₃ + x₁² + [2]x₁·x₂   
POL(u32(x₁, x₂, x₃, x₄)) = [1] + x₁ + x₂ + [2]x₃ + x₄ + x₃·x₄ + x₂·x₄ + [2]x₁·x₄ + [2]x₁² + [2]x₁·x₂ + [2]x₁·x₃   
POL(u41(x₁, x₂)) = [2]x₁·x₂ + [2]x₁²   
POL(u42(x₁, x₂, x₃)) = [1] + [2]x₁ + x₂ + [2]x₃ + x₃² + x₂·x₃ + [2]x₁·x₃ + [2]x₁² + x₁·x₂   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:none
K tuples:

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(19) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(20) BOUNDS(1, 1)