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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsProof (⇔, 11 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 68 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(S(x)) → even(x)
even(S(x)) → odd(x)
odd(0) → 0
even(0) → S(0)

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[odd_1|0, even_1|0, even_1|1, 0|1, odd_1|1]
1→3[S_1|1]
2→2[S_1|0, 0|0]
3→2[0|1]

(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(S(z0)) → even(z0)
odd(0) → 0
even(S(z0)) → odd(z0)
even(0) → S(0)
Tuples:

ODD(S(z0)) → c(EVEN(z0))
ODD(0) → c1
EVEN(S(z0)) → c2(ODD(z0))
EVEN(0) → c3
S tuples:

ODD(S(z0)) → c(EVEN(z0))
ODD(0) → c1
EVEN(S(z0)) → c2(ODD(z0))
EVEN(0) → c3
K tuples:none
Defined Rule Symbols:

odd, even

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

EVEN(0) → c3
ODD(0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

odd(S(z0)) → even(z0)
odd(0) → 0
even(S(z0)) → odd(z0)
even(0) → S(0)
Tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))
S tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))
K tuples:none
Defined Rule Symbols:

odd, even

Defined Pair Symbols:

ODD, EVEN

Compound Symbols:

c, c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

odd(S(z0)) → even(z0)
odd(0) → 0
even(S(z0)) → odd(z0)
even(0) → S(0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))
S tuples:

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

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

POL(EVEN(x1)) = x1   
POL(ODD(x1)) = x1   
POL(S(x1)) = [1] + x1   
POL(c(x1)) = x1   
POL(c2(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

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

ODD(S(z0)) → c(EVEN(z0))
EVEN(S(z0)) → c2(ODD(z0))
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)