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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 CdtProblem
↳15 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳16 CdtProblem
↳17 CdtKnowledgeProof (⇔, 0 ms)
↳18 BOUNDS(O(1), O(1))

(0) Obligation:

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

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, is_empty(l1))
ifappend(l1, l2, true) → l2
ifappend(l1, l2, false) → cons(hd(l1), append(tl(l1), l2))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, is_empty(z0)), IS_EMPTY(z0))
IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
S tuples:

APPEND(z0, z1) → c4(IFAPPEND(z0, z1, is_empty(z0)), IS_EMPTY(z0))
IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

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

Use narrowing to replace APPEND(z0, z1) → c4(IFAPPEND(z0, z1, is_empty(z0)), IS_EMPTY(z0)) by

APPEND(nil, x1) → c4(IFAPPEND(nil, x1, true), IS_EMPTY(nil))
APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
APPEND(x0, x1) → c4

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
APPEND(nil, x1) → c4(IFAPPEND(nil, x1, true), IS_EMPTY(nil))
APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
APPEND(x0, x1) → c4
S tuples:

IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
APPEND(nil, x1) → c4(IFAPPEND(nil, x1, true), IS_EMPTY(nil))
APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
APPEND(x0, x1) → c4
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

IFAPPEND, APPEND

Compound Symbols:

c6, c4, c4

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

Removed 2 trailing nodes:

APPEND(x0, x1) → c4
APPEND(nil, x1) → c4(IFAPPEND(nil, x1, true), IS_EMPTY(nil))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
S tuples:

IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

IFAPPEND, APPEND

Compound Symbols:

c6, c4

(7) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0)) by

IFAPPEND(x0, x1, false) → c6(APPEND(tl(x0), x1))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
IFAPPEND(x0, x1, false) → c6(APPEND(tl(x0), x1))
S tuples:

APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1)))
IFAPPEND(x0, x1, false) → c6(APPEND(tl(x0), x1))
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false), IS_EMPTY(cons(z0, z1))) by

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

IFAPPEND(x0, x1, false) → c6(APPEND(tl(x0), x1))
APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
S tuples:

IFAPPEND(x0, x1, false) → c6(APPEND(tl(x0), x1))
APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

IFAPPEND, APPEND

Compound Symbols:

c6, c4

(11) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace IFAPPEND(x0, x1, false) → c6(APPEND(tl(x0), x1)) by

IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
IFAPPEND(x0, x1, false) → c6

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
IFAPPEND(x0, x1, false) → c6
S tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
IFAPPEND(x0, x1, false) → c6
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6, c6

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

Removed 1 trailing nodes:

IFAPPEND(x0, x1, false) → c6

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
S tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
K tuples:none
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(15) 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.

IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
We considered the (Usable) Rules:none
And the Tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(APPEND(x1, x2)) = x1   
POL(IFAPPEND(x1, x2, x3)) = x1   
POL(c4(x1)) = x1   
POL(c6(x1)) = x1   
POL(cons(x1, x2)) = [4] + x1 + x2   
POL(false) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false
hd(cons(z0, z1)) → z0
tl(cons(z0, z1)) → z1
append(z0, z1) → ifappend(z0, z1, is_empty(z0))
ifappend(z0, z1, true) → z1
ifappend(z0, z1, false) → cons(hd(z0), append(tl(z0), z1))
Tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
S tuples:

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
K tuples:

IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
Defined Rule Symbols:

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(17) CdtKnowledgeProof (EQUIVALENT transformation)

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

APPEND(cons(x0, x1), x2) → c4(IFAPPEND(cons(x0, x1), x2, false))
IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
Now S is empty

(18) BOUNDS(O(1), O(1))