The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^3)).

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
↳4 CdtProblem
↳5 CdtUsableRulesProof (⇔, 0 ms)
↳6 CdtProblem
↳7 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 122 ms)
↳8 CdtProblem
↳9 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))), 126 ms)
↳10 CdtProblem
↳11 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))), 199 ms)
↳12 CdtProblem
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 BOUNDS(O(1), O(1))

(0) Obligation:

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

shuffle(Cons(x, xs)) → Cons(x, shuffle(reverse(xs)))
reverse(Cons(x, xs)) → append(reverse(xs), Cons(x, Nil))
append(Cons(x, xs), ys) → Cons(x, append(xs, ys))
shuffle(Nil) → Nil
reverse(Nil) → Nil
append(Nil, ys) → ys
goal(xs) → shuffle(xs)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

shuffle(Cons(z0, z1)) → Cons(z0, shuffle(reverse(z1)))
shuffle(Nil) → Nil
reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
goal(z0) → shuffle(z0)
Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
SHUFFLE(Nil) → c1
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
REVERSE(Nil) → c3
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
APPEND(Nil, z0) → c5
GOAL(z0) → c6(SHUFFLE(z0))
S tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
SHUFFLE(Nil) → c1
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
REVERSE(Nil) → c3
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
APPEND(Nil, z0) → c5
GOAL(z0) → c6(SHUFFLE(z0))
K tuples:none
Defined Rule Symbols:

shuffle, reverse, append, goal

Defined Pair Symbols:

SHUFFLE, REVERSE, APPEND, GOAL

Compound Symbols:

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

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0) → c6(SHUFFLE(z0))
Removed 3 trailing nodes:

REVERSE(Nil) → c3
SHUFFLE(Nil) → c1
APPEND(Nil, z0) → c5

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

shuffle(Cons(z0, z1)) → Cons(z0, shuffle(reverse(z1)))
shuffle(Nil) → Nil
reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
goal(z0) → shuffle(z0)
Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
S tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
K tuples:none
Defined Rule Symbols:

shuffle, reverse, append, goal

Defined Pair Symbols:

SHUFFLE, REVERSE, APPEND

Compound Symbols:

c, c2, c4

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

shuffle(Cons(z0, z1)) → Cons(z0, shuffle(reverse(z1)))
shuffle(Nil) → Nil
goal(z0) → shuffle(z0)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
S tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
K tuples:none
Defined Rule Symbols:

reverse, append

Defined Pair Symbols:

SHUFFLE, REVERSE, APPEND

Compound Symbols:

c, c2, c4

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

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

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
We considered the (Usable) Rules:

reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
append(Nil, z0) → z0
And the Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(APPEND(x1, x2)) = 0   
POL(Cons(x1, x2)) = [1] + x2   
POL(Nil) = 0   
POL(REVERSE(x1)) = [1]   
POL(SHUFFLE(x1)) = [2]x1   
POL(append(x1, x2)) = x1 + x2   
POL(c(x1, x2)) = x1 + x2   
POL(c2(x1, x2)) = x1 + x2   
POL(c4(x1)) = x1   
POL(reverse(x1)) = x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
S tuples:

REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
K tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
Defined Rule Symbols:

reverse, append

Defined Pair Symbols:

SHUFFLE, REVERSE, APPEND

Compound Symbols:

c, c2, c4

(9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
We considered the (Usable) Rules:

reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
append(Nil, z0) → z0
And the Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(APPEND(x1, x2)) = 0   
POL(Cons(x1, x2)) = [1] + x2   
POL(Nil) = 0   
POL(REVERSE(x1)) = [1] + x1   
POL(SHUFFLE(x1)) = x12   
POL(append(x1, x2)) = x1 + x2   
POL(c(x1, x2)) = x1 + x2   
POL(c2(x1, x2)) = x1 + x2   
POL(c4(x1)) = x1   
POL(reverse(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
S tuples:

APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
K tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
Defined Rule Symbols:

reverse, append

Defined Pair Symbols:

SHUFFLE, REVERSE, APPEND

Compound Symbols:

c, c2, c4

(11) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))) transformation)

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

APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
We considered the (Usable) Rules:

reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
append(Nil, z0) → z0
And the Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(APPEND(x1, x2)) = x1 + x2 + x23   
POL(Cons(x1, x2)) = [1] + x2   
POL(Nil) = 0   
POL(REVERSE(x1)) = [1] + x1 + x12   
POL(SHUFFLE(x1)) = x1 + x13   
POL(append(x1, x2)) = x1 + x2   
POL(c(x1, x2)) = x1 + x2   
POL(c2(x1, x2)) = x1 + x2   
POL(c4(x1)) = x1   
POL(reverse(x1)) = x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

reverse(Cons(z0, z1)) → append(reverse(z1), Cons(z0, Nil))
reverse(Nil) → Nil
append(Cons(z0, z1), z2) → Cons(z0, append(z1, z2))
append(Nil, z0) → z0
Tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
S tuples:none
K tuples:

SHUFFLE(Cons(z0, z1)) → c(SHUFFLE(reverse(z1)), REVERSE(z1))
REVERSE(Cons(z0, z1)) → c2(APPEND(reverse(z1), Cons(z0, Nil)), REVERSE(z1))
APPEND(Cons(z0, z1), z2) → c4(APPEND(z1, z2))
Defined Rule Symbols:

reverse, append

Defined Pair Symbols:

SHUFFLE, REVERSE, APPEND

Compound Symbols:

c, c2, c4

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

The set S is empty

(14) BOUNDS(O(1), O(1))