### (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