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

(0) 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:

mul0(Cons(x, xs), y) → add0(mul0(xs, y), y)
add0(Cons(x, xs), y) → add0(xs, Cons(S, y))
mul0(Nil, y) → Nil
add0(Nil, y) → y
goal(xs, ys) → mul0(xs, ys)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
goal(z0, z1) → mul0(z0, z1)

Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
MUL0(Nil, z0) → c1
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
ADD0(Nil, z0) → c3
GOAL(z0, z1) → c4(MUL0(z0, z1))

S tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
MUL0(Nil, z0) → c1
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
ADD0(Nil, z0) → c3
GOAL(z0, z1) → c4(MUL0(z0, z1))

K tuples:none
Defined Rule Symbols:

mul0, add0, goal

Defined Pair Symbols:

MUL0, ADD0, GOAL

Compound Symbols:

c, c1, c2, c3, c4

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c4(MUL0(z0, z1))

Removed 2 trailing nodes:

ADD0(Nil, z0) → c3
MUL0(Nil, z0) → c1

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
goal(z0, z1) → mul0(z0, z1)

Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

S tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

K tuples:none
Defined Rule Symbols:

mul0, add0, goal

Defined Pair Symbols:

MUL0, ADD0

Compound Symbols:

c, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

goal(z0, z1) → mul0(z0, z1)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0

Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

S tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

K tuples:none
Defined Rule Symbols:

mul0, add0

Defined Pair Symbols:

MUL0, ADD0

Compound Symbols:

c, c2

(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.

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))

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

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

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

POL(ADD0(x₁, x₂)) = 0   
POL(Cons(x₁, x₂)) = [1] + x₂   
POL(MUL0(x₁, x₂)) = x₁   
POL(Nil) = 0   
POL(S) = 0   
POL(add0(x₁, x₂)) = 0   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(mul0(x₁, x₂)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0

Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

S tuples:

ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

K tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))

Defined Rule Symbols:

mul0, add0

Defined Pair Symbols:

MUL0, ADD0

Compound Symbols:

c, c2

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

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

ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

We considered the (Usable) Rules:

mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)

And the Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

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

POL(ADD0(x₁, x₂)) = [1] + x₁   
POL(Cons(x₁, x₂)) = [1] + x₂   
POL(MUL0(x₁, x₂)) = x₁³ + x₁²·x₂   
POL(Nil) = 0   
POL(S) = 0   
POL(add0(x₁, x₂)) = x₁ + x₂   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(mul0(x₁, x₂)) = x₁·x₂   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0

Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

S tuples:none
K tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))

Defined Rule Symbols:

mul0, add0

Defined Pair Symbols:

MUL0, ADD0

Compound Symbols:

c, c2

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

The set S is empty

(12) BOUNDS(1, 1)