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

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

q0(a(x1)) → x(q1(x1))
q1(a(x1)) → a(q1(x1))
q1(y(x1)) → y(q1(x1))
a(q1(b(x1))) → q2(a(y(x1)))
a(q2(a(x1))) → q2(a(a(x1)))
a(q2(y(x1))) → q2(a(y(x1)))
y(q1(b(x1))) → q2(y(y(x1)))
y(q2(a(x1))) → q2(y(a(x1)))
y(q2(y(x1))) → q2(y(y(x1)))
q2(x(x1)) → x(q0(x1))
q0(y(x1)) → y(q3(x1))
q3(y(x1)) → y(q3(x1))
q3(bl(x1)) → bl(q4(x1))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
A(q2(a(z0))) → c5(Q2(a(a(z0))), A(a(z0)), A(z0))
A(q2(y(z0))) → c6(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Y(q2(a(z0))) → c8(Q2(y(a(z0))), Y(a(z0)), A(z0))
Y(q2(y(z0))) → c9(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))

S tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
A(q2(a(z0))) → c5(Q2(a(a(z0))), A(a(z0)), A(z0))
A(q2(y(z0))) → c6(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Y(q2(a(z0))) → c8(Q2(y(a(z0))), Y(a(z0)), A(z0))
Y(q2(y(z0))) → c9(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))

K tuples:none
Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11

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

Removed 8 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:none
Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = 0   
POL(Q0(x₁)) = [4]   
POL(Q1(x₁)) = 0   
POL(Q2(x₁)) = [4]x₁   
POL(Q3(x₁)) = 0   
POL(Y(x₁)) = 0   
POL(a(x₁)) = 0   
POL(b(x₁)) = [4]   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [5] + [2]x₁   
POL(q1(x₁)) = [2]   
POL(q2(x₁)) = [4]x₁   
POL(q3(x₁)) = [3] + [5]x₁   
POL(q4(x₁)) = [3]   
POL(x(x₁)) = [1]   
POL(y(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q0(a(z0)) → c(Q1(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

Q2(x(z0)) → c10(Q0(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = 0   
POL(Q0(x₁)) = [2]   
POL(Q1(x₁)) = 0   
POL(Q2(x₁)) = [4]x₁   
POL(Q3(x₁)) = [1]   
POL(Y(x₁)) = 0   
POL(a(x₁)) = 0   
POL(b(x₁)) = 0   
POL(bl(x₁)) = [3]   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [4]   
POL(q1(x₁)) = 0   
POL(q2(x₁)) = [4]x₁   
POL(q3(x₁)) = [4] + [5]x₁   
POL(q4(x₁)) = [3]   
POL(x(x₁)) = [1]   
POL(y(x₁)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q0(a(z0)) → c(Q1(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

Q0(a(z0)) → c(Q1(z0))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = [2]x₁²   
POL(Q0(x₁)) = [2] + x₁   
POL(Q1(x₁)) = [3] + [2]x₁²   
POL(Q2(x₁)) = [1] + x₁   
POL(Q3(x₁)) = [2] + x₁   
POL(Y(x₁)) = [2] + [2]x₁   
POL(a(x₁)) = [2] + [3]x₁ + [2]x₁²   
POL(b(x₁)) = [3] + x₁   
POL(bl(x₁)) = [2]   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [2] + [2]x₁   
POL(q1(x₁)) = [3]x₁ + x₁²   
POL(q2(x₁)) = [1] + [2]x₁   
POL(q3(x₁)) = x₁   
POL(q4(x₁)) = 0   
POL(x(x₁)) = [3] + x₁   
POL(y(x₁)) = [2] + [3]x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = x₁²   
POL(Q0(x₁)) = x₁   
POL(Q1(x₁)) = [2]x₁²   
POL(Q2(x₁)) = [2]x₁   
POL(Q3(x₁)) = x₁   
POL(Y(x₁)) = x₁   
POL(a(x₁)) = [3]x₁ + [2]x₁²   
POL(b(x₁)) = [2] + x₁   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = x₁   
POL(q1(x₁)) = [3]x₁ + [2]x₁²   
POL(q2(x₁)) = x₁   
POL(q3(x₁)) = x₁   
POL(q4(x₁)) = 0   
POL(x(x₁)) = x₁   
POL(y(x₁)) = [2] + [2]x₁   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = x₁²   
POL(Q0(x₁)) = x₁   
POL(Q1(x₁)) = [2]x₁ + x₁²   
POL(Q2(x₁)) = x₁   
POL(Q3(x₁)) = 0   
POL(Y(x₁)) = x₁   
POL(a(x₁)) = [2]x₁ + x₁²   
POL(b(x₁)) = [3] + x₁   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [2] + [2]x₁   
POL(q1(x₁)) = x₁ + x₁²   
POL(q2(x₁)) = [2] + [2]x₁   
POL(q3(x₁)) = 0   
POL(q4(x₁)) = 0   
POL(x(x₁)) = x₁   
POL(y(x₁)) = [2]x₁   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(17) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = [1] + x₁²   
POL(Q0(x₁)) = [2] + x₁   
POL(Q1(x₁)) = [3] + [2]x₁²   
POL(Q2(x₁)) = [2]x₁   
POL(Q3(x₁)) = [1]   
POL(Y(x₁)) = [2]x₁   
POL(a(x₁)) = [2] + [2]x₁²   
POL(b(x₁)) = [2] + x₁   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [3] + x₁   
POL(q1(x₁)) = [2]x₁²   
POL(q2(x₁)) = [2] + [2]x₁   
POL(q3(x₁)) = 0   
POL(q4(x₁)) = 0   
POL(x(x₁)) = [3] + x₁   
POL(y(x₁)) = [2]x₁   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(19) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = x₁²   
POL(Q0(x₁)) = [3] + x₁   
POL(Q1(x₁)) = [1] + [2]x₁²   
POL(Q2(x₁)) = [1] + [2]x₁   
POL(Q3(x₁)) = 0   
POL(Y(x₁)) = [2]x₁   
POL(a(x₁)) = [1] + [2]x₁²   
POL(b(x₁)) = [2] + x₁   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [3] + x₁   
POL(q1(x₁)) = [2]x₁²   
POL(q2(x₁)) = [1] + [2]x₁   
POL(q3(x₁)) = 0   
POL(q4(x₁)) = 0   
POL(x(x₁)) = [3] + x₁   
POL(y(x₁)) = [2]x₁   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(21) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

A(q2(a(z0))) → c5(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = x₁²   
POL(Q0(x₁)) = [1] + [2]x₁   
POL(Q1(x₁)) = x₁ + x₁²   
POL(Q2(x₁)) = [2] + [2]x₁   
POL(Q3(x₁)) = 0   
POL(Y(x₁)) = x₁   
POL(a(x₁)) = [2]x₁ + [2]x₁²   
POL(b(x₁)) = [3] + x₁   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [1] + x₁   
POL(q1(x₁)) = x₁ + [2]x₁²   
POL(q2(x₁)) = [1] + [2]x₁   
POL(q3(x₁)) = 0   
POL(q4(x₁)) = 0   
POL(x(x₁)) = [1] + x₁   
POL(y(x₁)) = [2]x₁   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:

Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))

K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
A(q2(a(z0))) → c5(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(23) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))

We considered the (Usable) Rules:

q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))

And the Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

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

POL(A(x₁)) = [3] + x₁²   
POL(Q0(x₁)) = [1] + x₁   
POL(Q1(x₁)) = [3] + x₁²   
POL(Q2(x₁)) = [2] + x₁   
POL(Q3(x₁)) = [3] + x₁   
POL(Y(x₁)) = x₁   
POL(a(x₁)) = [3] + [3]x₁ + [2]x₁²   
POL(b(x₁)) = [2] + x₁   
POL(bl(x₁)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c10(x₁)) = x₁   
POL(c11(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c5(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c8(x₁)) = x₁   
POL(c9(x₁)) = x₁   
POL(q0(x₁)) = [1] + [2]x₁   
POL(q1(x₁)) = [3]x₁ + x₁²   
POL(q2(x₁)) = [2] + [2]x₁   
POL(q3(x₁)) = x₁   
POL(q4(x₁)) = 0   
POL(x(x₁)) = x₁   
POL(y(x₁)) = [2] + [2]x₁   

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

q0(a(z0)) → x(q1(z0))
q0(y(z0)) → y(q3(z0))
q1(a(z0)) → a(q1(z0))
q1(y(z0)) → y(q1(z0))
a(q1(b(z0))) → q2(a(y(z0)))
a(q2(a(z0))) → q2(a(a(z0)))
a(q2(y(z0))) → q2(a(y(z0)))
y(q1(b(z0))) → q2(y(y(z0)))
y(q2(a(z0))) → q2(y(a(z0)))
y(q2(y(z0))) → q2(y(y(z0)))
q2(x(z0)) → x(q0(z0))
q3(y(z0)) → y(q3(z0))
q3(bl(z0)) → bl(q4(z0))

Tuples:

Q0(a(z0)) → c(Q1(z0))
Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
Q2(x(z0)) → c10(Q0(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))
A(q2(a(z0))) → c5(A(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Y(q2(y(z0))) → c9(Y(z0))

S tuples:none
K tuples:

Q0(y(z0)) → c1(Y(q3(z0)), Q3(z0))
Q2(x(z0)) → c10(Q0(z0))
Q0(a(z0)) → c(Q1(z0))
A(q1(b(z0))) → c4(Q2(a(y(z0))), A(y(z0)), Y(z0))
Q1(y(z0)) → c3(Y(q1(z0)), Q1(z0))
A(q2(y(z0))) → c6(Y(z0))
Y(q2(a(z0))) → c8(A(z0))
Q1(a(z0)) → c2(A(q1(z0)), Q1(z0))
Y(q1(b(z0))) → c7(Q2(y(y(z0))), Y(y(z0)), Y(z0))
A(q2(a(z0))) → c5(A(z0))
Y(q2(y(z0))) → c9(Y(z0))
Q3(y(z0)) → c11(Y(q3(z0)), Q3(z0))

Defined Rule Symbols:

q0, q1, a, y, q2, q3

Defined Pair Symbols:

Q0, Q1, A, Y, Q2, Q3

Compound Symbols:

c, c1, c2, c3, c4, c7, c10, c11, c5, c6, c8, c9

(25) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(26) BOUNDS(O(1), O(1))