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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
*(s(x), [])
*(p(x), [])

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

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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(0) → c3
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(0, z0) → c6
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(0) → c3
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(0, z0) → c6
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

K tuples:none
Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8

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

Removed 3 trailing nodes:

+'(0, z0) → c
*'(0, z0) → c6
MINUS(0) → c3

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

K tuples:none
Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c1, c2, c4, c5, c7, c8

(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.

*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

POL(*(x₁, x₂)) = 0   
POL(*'(x₁, x₂)) = [2]x₁   
POL(+(x₁, x₂)) = 0   
POL(+'(x₁, x₂)) = 0   
POL(0) = 0   
POL(MINUS(x₁)) = [3]   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(minus(x₁)) = 0   
POL(p(x₁)) = [2] + x₁   
POL(s(x₁)) = [3] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))

K tuples:

*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c1, c2, c4, c5, c7, c8

(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.

MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))

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

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

POL(*(x₁, x₂)) = [2]x₂² + [2]x₁²   
POL(*'(x₁, x₂)) = [2]x₁·x₂   
POL(+(x₁, x₂)) = [1] + x₁ + [2]x₂ + [2]x₂² + x₁²   
POL(+'(x₁, x₂)) = 0   
POL(0) = [1]   
POL(MINUS(x₁)) = [2]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(minus(x₁)) = [1] + x₁   
POL(p(x₁)) = [1] + x₁   
POL(s(x₁)) = [2] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))

K tuples:

*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))

Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c1, c2, c4, c5, c7, c8

(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.

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))

We considered the (Usable) Rules:

+(s(z0), z1) → s(+(z0, z1))
*(p(z0), z1) → +(*(z0, z1), minus(z1))
minus(0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
minus(s(z0)) → p(minus(z0))
+(p(z0), z1) → p(+(z0, z1))
+(0, z0) → z0

And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

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

POL(*(x₁, x₂)) = x₁·x₂   
POL(*'(x₁, x₂)) = x₁²·x₂   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(+'(x₁, x₂)) = x₁   
POL(0) = 0   
POL(MINUS(x₁)) = 0   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(minus(x₁)) = x₁   
POL(p(x₁)) = [1] + x₁   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
+(p(z0), z1) → p(+(z0, z1))
minus(0) → 0
minus(s(z0)) → p(minus(z0))
minus(p(z0)) → s(minus(z0))
*(0, z0) → 0
*(s(z0), z1) → +(*(z0, z1), z1)
*(p(z0), z1) → +(*(z0, z1), minus(z1))

Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))

S tuples:none
K tuples:

*'(s(z0), z1) → c7(+'(*(z0, z1), z1), *'(z0, z1))
*'(p(z0), z1) → c8(+'(*(z0, z1), minus(z1)), *'(z0, z1), MINUS(z1))
MINUS(s(z0)) → c4(MINUS(z0))
MINUS(p(z0)) → c5(MINUS(z0))
+'(s(z0), z1) → c1(+'(z0, z1))
+'(p(z0), z1) → c2(+'(z0, z1))

Defined Rule Symbols:

+, minus, *

Defined Pair Symbols:

+', MINUS, *'

Compound Symbols:

c1, c2, c4, c5, c7, c8

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

The set S is empty

(14) BOUNDS(1, 1)