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

0 CpxTRS
↳1 RcToIrcProof (BOTH BOUNDS(ID, ID), 22 ms)
↳2 CpxTRS
↳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)), 75 ms)
↳10 CdtProblem
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 BOUNDS(1, 1)

(0) Obligation:

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


The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

(2) 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:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, x)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
sum1(0) → 0
sum1(s(z0)) → s(+(sum1(z0), +(z0, z0)))
Tuples:

SUM(0) → c
SUM(s(z0)) → c1(SUM(z0))
SUM1(0) → c2
SUM1(s(z0)) → c3(SUM1(z0))
S tuples:

SUM(0) → c
SUM(s(z0)) → c1(SUM(z0))
SUM1(0) → c2
SUM1(s(z0)) → c3(SUM1(z0))
K tuples:none
Defined Rule Symbols:

sum, sum1

Defined Pair Symbols:

SUM, SUM1

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

SUM(0) → c
SUM1(0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
sum1(0) → 0
sum1(s(z0)) → s(+(sum1(z0), +(z0, z0)))
Tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
S tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
K tuples:none
Defined Rule Symbols:

sum, sum1

Defined Pair Symbols:

SUM, SUM1

Compound Symbols:

c1, c3

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
sum1(0) → 0
sum1(s(z0)) → s(+(sum1(z0), +(z0, z0)))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
S tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

SUM, SUM1

Compound Symbols:

c1, c3

(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.

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
We considered the (Usable) Rules:none
And the Tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(SUM(x1)) = x1   
POL(SUM1(x1)) = x1   
POL(c1(x1)) = x1   
POL(c3(x1)) = x1   
POL(s(x1)) = [2] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
S tuples:none
K tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

SUM, SUM1

Compound Symbols:

c1, c3

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

The set S is empty

(12) BOUNDS(1, 1)