The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

0 CpxTRS
↳1 RcToIrcProof (BOTH BOUNDS(ID, ID), 0 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)), 105 ms)
↳10 CdtProblem
↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 0 ms)
↳12 CdtProblem
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 BOUNDS(1, 1)

(0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

average(s(x), y) → average(x, s(y))
average(x, s(s(s(y)))) → s(average(s(x), y))
average(0, 0) → 0
average(0, s(0)) → 0
average(0, s(s(0))) → s(0)

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^2).


The TRS R consists of the following rules:

average(s(x), y) → average(x, s(y))
average(x, s(s(s(y)))) → s(average(s(x), y))
average(0, 0) → 0
average(0, s(0)) → 0
average(0, s(s(0))) → s(0)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

average(s(z0), z1) → average(z0, s(z1))
average(z0, s(s(s(z1)))) → s(average(s(z0), z1))
average(0, 0) → 0
average(0, s(0)) → 0
average(0, s(s(0))) → s(0)
Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
AVERAGE(0, 0) → c2
AVERAGE(0, s(0)) → c3
AVERAGE(0, s(s(0))) → c4
S tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
AVERAGE(0, 0) → c2
AVERAGE(0, s(0)) → c3
AVERAGE(0, s(s(0))) → c4
K tuples:none
Defined Rule Symbols:

average

Defined Pair Symbols:

AVERAGE

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 3 trailing nodes:

AVERAGE(0, 0) → c2
AVERAGE(0, s(0)) → c3
AVERAGE(0, s(s(0))) → c4

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

average(s(z0), z1) → average(z0, s(z1))
average(z0, s(s(s(z1)))) → s(average(s(z0), z1))
average(0, 0) → 0
average(0, s(0)) → 0
average(0, s(s(0))) → s(0)
Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
S tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
K tuples:none
Defined Rule Symbols:

average

Defined Pair Symbols:

AVERAGE

Compound Symbols:

c, c1

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

average(s(z0), z1) → average(z0, s(z1))
average(z0, s(s(s(z1)))) → s(average(s(z0), z1))
average(0, 0) → 0
average(0, s(0)) → 0
average(0, s(s(0))) → s(0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
S tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

AVERAGE

Compound Symbols:

c, c1

(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.

AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(AVERAGE(x1, x2)) = x1 + x2   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(s(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
S tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
K tuples:

AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
Defined Rule Symbols:none

Defined Pair Symbols:

AVERAGE

Compound Symbols:

c, c1

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
We considered the (Usable) Rules:none
And the Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(AVERAGE(x1, x2)) = x22 + [2]x1·x2 + [2]x12   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(s(x1)) = [2] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
S tuples:none
K tuples:

AVERAGE(z0, s(s(s(z1)))) → c1(AVERAGE(s(z0), z1))
AVERAGE(s(z0), z1) → c(AVERAGE(z0, s(z1)))
Defined Rule Symbols:none

Defined Pair Symbols:

AVERAGE

Compound Symbols:

c, c1

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

The set S is empty

(14) BOUNDS(1, 1)