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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtUsableRulesProof (⇔, 0 ms)
↳14 CdtProblem
↳15 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳16 CdtProblem
↳17 CdtUsableRulesProof (⇔, 0 ms)
↳18 CdtProblem
↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 12 ms)
↳20 CdtProblem
↳21 CdtKnowledgeProof (⇔, 0 ms)
↳22 BOUNDS(1, 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 Cpx (relative) TRS 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:

IS_EMPTY(nil) → c
IS_EMPTY(cons(z0, z1)) → c1
HD(cons(z0, z1)) → c2
TL(cons(z0, z1)) → c3
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, is_empty(z0)), IS_EMPTY(z0))
IFAPPEND(z0, z1, true) → c5
IFAPPEND(z0, z1, false) → c6(HD(z0), APPEND(tl(z0), z1), TL(z0))
S tuples:

IS_EMPTY(nil) → c
IS_EMPTY(cons(z0, z1)) → c1
HD(cons(z0, z1)) → c2
TL(cons(z0, z1)) → c3
APPEND(z0, z1) → c4(IFAPPEND(z0, z1, is_empty(z0)), IS_EMPTY(z0))
IFAPPEND(z0, z1, true) → c5
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:

IS_EMPTY, HD, TL, APPEND, IFAPPEND

Compound Symbols:

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

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

Removed 5 trailing nodes:

IS_EMPTY(cons(z0, z1)) → c1
IFAPPEND(z0, z1, true) → c5
TL(cons(z0, z1)) → c3
HD(cons(z0, z1)) → c2
IS_EMPTY(nil) → c

(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:

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

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

Removed 3 trailing tuple parts

(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:

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

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

is_empty, hd, tl, append, ifappend

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

hd(cons(z0, z1)) → z0
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))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

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

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

is_empty, tl

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

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

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

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

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

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

is_empty, tl

Defined Pair Symbols:

IFAPPEND, APPEND

Compound Symbols:

c6, c4

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

Removed 1 trailing nodes:

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

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

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

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

is_empty, tl

Defined Pair Symbols:

IFAPPEND, APPEND

Compound Symbols:

c6, c4

(13) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

is_empty(nil) → true
is_empty(cons(z0, z1)) → false

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

tl(cons(z0, z1)) → z1
Tuples:

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

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

tl

Defined Pair Symbols:

IFAPPEND, APPEND

Compound Symbols:

c6, c4

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

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

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

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

tl(cons(z0, z1)) → z1
Tuples:

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

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

tl

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(17) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

tl(cons(z0, z1)) → z1

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

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

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

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(19) CdtRuleRemovalProof (UPPER BOUND(ADD(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(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, 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   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

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

APPEND(cons(z0, z1), x1) → c4(IFAPPEND(cons(z0, z1), x1, false))
K tuples:

IFAPPEND(cons(z0, z1), x1, false) → c6(APPEND(z1, x1))
Defined Rule Symbols:none

Defined Pair Symbols:

APPEND, IFAPPEND

Compound Symbols:

c4, c6

(21) CdtKnowledgeProof (EQUIVALENT transformation)

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

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

(22) BOUNDS(1, 1)