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

(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:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
sqr(s([]))

The defined contexts are:
+([], x1)
+(x0, [])
+([], s(x1))
+(x0, s([]))

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, 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:

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(0) → c
SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(0) → c3
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, 0) → c5
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

SQR(0) → c
SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(0) → c3
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, 0) → c5
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

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

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

Removed 3 trailing nodes:

SQR(0) → c
DOUBLE(0) → c3
+'(z0, 0) → c5

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))

We considered the (Usable) Rules:none
And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x₁, x₂)) = 0   
POL(+'(x₁, x₂)) = 0   
POL(0) = 0   
POL(DOUBLE(x₁)) = 0   
POL(SQR(x₁)) = x₁   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(double(x₁)) = 0   
POL(s(x₁)) = [1] + x₁   
POL(sqr(x₁)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))

Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

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

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

DOUBLE(s(z0)) → c4(DOUBLE(z0))

We considered the (Usable) Rules:none
And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x₁, x₂)) = [2]x₁ + x₂   
POL(+'(x₁, x₂)) = [2]   
POL(0) = [1]   
POL(DOUBLE(x₁)) = [2] + [2]x₁   
POL(SQR(x₁)) = x₁²   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(double(x₁)) = 0   
POL(s(x₁)) = [2] + x₁   
POL(sqr(x₁)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))

Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

(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.

+'(z0, s(z1)) → c6(+'(z0, z1))

We considered the (Usable) Rules:

double(0) → 0
double(s(z0)) → s(s(double(z0)))

And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x₁, x₂)) = [2] + x₁ + [2]x₂ + x₂² + x₁²   
POL(+'(x₁, x₂)) = x₂   
POL(0) = 0   
POL(DOUBLE(x₁)) = x₁   
POL(SQR(x₁)) = x₁ + x₁²   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(double(x₁)) = [2]x₁   
POL(s(x₁)) = [2] + x₁   
POL(sqr(x₁)) = [2]   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:none
K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

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

The set S is empty

(14) BOUNDS(1, 1)