The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^3).

(0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^3).


The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
times([], s(y))

The defined contexts are:
plus([], x1)

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^3).


The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

TIMES(z0, 0) → c
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, 0) → c2
PLUS(0, z0) → c3
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

S tuples:

TIMES(z0, 0) → c
TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, 0) → c2
PLUS(0, z0) → c3
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 3 trailing nodes:

TIMES(z0, 0) → c
PLUS(0, z0) → c3
PLUS(z0, 0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

S tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

(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.

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))

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

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

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

POL(0) = 0   
POL(PLUS(x₁, x₂)) = [3]   
POL(TIMES(x₁, x₂)) = [2]x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = [2] + x₁   
POL(times(x₁, x₂)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

S tuples:

PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

K tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))

Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

(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.

PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

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

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

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

POL(0) = 0   
POL(PLUS(x₁, x₂)) = [2]x₂   
POL(TIMES(x₁, x₂)) = x₂ + x₁·x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(plus(x₁, x₂)) = [2]x₁·x₂   
POL(s(x₁)) = [2] + x₁   
POL(times(x₁, x₂)) = [1]   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

S tuples:

PLUS(s(z0), z1) → c5(PLUS(z0, z1))

K tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^3)) transformation)

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

PLUS(s(z0), z1) → c5(PLUS(z0, z1))

We considered the (Usable) Rules:

times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
times(z0, 0) → 0

And the Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

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

POL(0) = 0   
POL(PLUS(x₁, x₂)) = x₁ + x₂ + x₂²   
POL(TIMES(x₁, x₂)) = x₁²·x₂ + x₁·x₂²   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = [1] + x₁   
POL(times(x₁, x₂)) = x₁·x₂   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(0, z0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
plus(s(z0), z1) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

S tuples:none
K tuples:

TIMES(z0, s(z1)) → c1(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))
PLUS(s(z0), z1) → c5(PLUS(z0, z1))

Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c1, c4, c5

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

The set S is empty

(14) BOUNDS(1, 1)