(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:

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

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

double(0) → 0
double(s(z0)) → s(s(double(z0)))
double(z0) → +(z0, z0)
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))
Tuples:

DOUBLE(0) → c
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, 0) → c3
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:

DOUBLE(0) → c
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, 0) → c3
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 2 trailing nodes:

DOUBLE(0) → c
+'(z0, 0) → c3

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0
double(s(z0)) → s(s(double(z0)))
double(z0) → +(z0, z0)
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

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

DOUBLE(z0) → c2(+'(z0, z0))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = 0
POL(DOUBLE(x1)) = [2]
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = 0

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:

DOUBLE(z0) → c2(+'(z0, z0))
Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

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

DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(s(z0), z1) → c5(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = [2]x1
POL(DOUBLE(x1)) = [2]x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:

+'(z0, s(z1)) → c4(+'(z0, z1))
K tuples:

DOUBLE(z0) → c2(+'(z0, z0))
DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(s(z0), z1) → c5(+'(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

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

+'(z0, s(z1)) → c4(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = x2
POL(DOUBLE(x1)) = x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:none
K tuples:

DOUBLE(z0) → c2(+'(z0, z0))
DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(s(z0), z1) → c5(+'(z0, z1))
+'(z0, s(z1)) → c4(+'(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

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

The set S is empty