(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)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(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)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(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)) → +(sqr(s(z0)), sum(z0))
sum(s(z0)) → +(*(s(z0), s(z0)), sum(z0))
sqr(z0) → *(z0, z0)
Tuples:

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

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

sum, sqr

Defined Pair Symbols:

SUM, SQR

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

SUM(0) → c
SQR(z0) → c3

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sqr(s(z0)), sum(z0))
sum(s(z0)) → +(*(s(z0), s(z0)), sum(z0))
sqr(z0) → *(z0, z0)
Tuples:

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))
S tuples:

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))
K tuples:none
Defined Rule Symbols:

sum, sqr

Defined Pair Symbols:

SUM

Compound Symbols:

c1, c2

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sqr(s(z0)), sum(z0))
sum(s(z0)) → +(*(s(z0), s(z0)), sum(z0))
sqr(z0) → *(z0, z0)
Tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
S tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
K tuples:none
Defined Rule Symbols:

sum, sqr

Defined Pair Symbols:

SUM

Compound Symbols:

c2, c1

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

sum(0) → 0
sum(s(z0)) → +(sqr(s(z0)), sum(z0))
sum(s(z0)) → +(*(s(z0), s(z0)), sum(z0))
sqr(z0) → *(z0, z0)

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
S tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

SUM

Compound Symbols:

c2, c1

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

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
We considered the (Usable) Rules:none
And the Tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(SUM(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
S tuples:none
K tuples:

SUM(s(z0)) → c2(SUM(z0))
SUM(s(z0)) → c1(SUM(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

SUM

Compound Symbols:

c2, c1

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

The set S is empty