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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 9 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 51 ms)
        ↳10 CdtProblem
        ↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳12 BOUNDS(1, 1)

(0) Obligation:

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


The TRS R consists of the following rules:

odd(Cons(x, xs)) → even(xs)
odd(Nil) → False
even(Cons(x, xs)) → odd(xs)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
even(Nil) → True
evenodd(x) → even(x)

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3, 4]
transitions:
Cons0(0, 0) → 0
Nil0() → 0
False0() → 0
True0() → 0
odd0(0) → 1
even0(0) → 2
notEmpty0(0) → 3
evenodd0(0) → 4
even1(0) → 1
False1() → 1
odd1(0) → 2
True1() → 3
False1() → 3
True1() → 2
even1(0) → 4
even1(0) → 2
False1() → 2
odd1(0) → 1
odd1(0) → 4
True1() → 1
True1() → 4
False1() → 4

(2) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(Cons(z0, z1)) → even(z1)
odd(Nil) → False
even(Cons(z0, z1)) → odd(z1)
even(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
evenodd(z0) → even(z0)
Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
ODD(Nil) → c1
EVEN(Cons(z0, z1)) → c2(ODD(z1))
EVEN(Nil) → c3
NOTEMPTY(Cons(z0, z1)) → c4
NOTEMPTY(Nil) → c5
EVENODD(z0) → c6(EVEN(z0))
S tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
ODD(Nil) → c1
EVEN(Cons(z0, z1)) → c2(ODD(z1))
EVEN(Nil) → c3
NOTEMPTY(Cons(z0, z1)) → c4
NOTEMPTY(Nil) → c5
EVENODD(z0) → c6(EVEN(z0))
K tuples:none
Defined Rule Symbols:

odd, even, notEmpty, evenodd

Defined Pair Symbols:

ODD, EVEN, NOTEMPTY, EVENODD

Compound Symbols:

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

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

EVENODD(z0) → c6(EVEN(z0))
Removed 4 trailing nodes:

NOTEMPTY(Cons(z0, z1)) → c4
ODD(Nil) → c1
NOTEMPTY(Nil) → c5
EVEN(Nil) → c3

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(Cons(z0, z1)) → even(z1)
odd(Nil) → False
even(Cons(z0, z1)) → odd(z1)
even(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
evenodd(z0) → even(z0)
Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
S tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
K tuples:none
Defined Rule Symbols:

odd, even, notEmpty, evenodd

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

odd(Cons(z0, z1)) → even(z1)
odd(Nil) → False
even(Cons(z0, z1)) → odd(z1)
even(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
evenodd(z0) → even(z0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
S tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(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.

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
We considered the (Usable) Rules:none
And the Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(Cons(x1, x2)) = [1] + x2   
POL(EVEN(x1)) = x1   
POL(ODD(x1)) = x1   
POL(c(x1)) = x1   
POL(c2(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
S tuples:none
K tuples:

ODD(Cons(z0, z1)) → c(EVEN(z1))
EVEN(Cons(z0, z1)) → c2(ODD(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

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

The set S is empty

(12) BOUNDS(1, 1)