YES(O(1), O(n^1)) 0.00/0.73 YES(O(1), O(n^1)) 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 0.00/0.75 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.75 0.00/0.75 0.00/0.75
0.00/0.75
0.00/0.75
0.00/0.75
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
0.00/0.75 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.75 
↳2 CdtProblem
0.00/0.75 
↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication)
0.00/0.75 
↳4 CdtProblem
0.00/0.75 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.75 
↳6 CdtProblem
0.00/0.75 
↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.75 
↳8 BOUNDS(O(1), O(1))
0.00/0.75
0.00/0.75 0.00/0.75
0.00/0.75
0.00/0.75

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

add0(x', Cons(x, xs)) → add0(Cons(Cons(Nil, Nil), x'), xs) 0.00/0.75
notEmpty(Cons(x, xs)) → True 0.00/0.75
notEmpty(Nil) → False 0.00/0.75
add0(x, Nil) → x 0.00/0.75
goal(x, y) → add0(x, y)

Rewrite Strategy: INNERMOST
0.00/0.75
0.00/0.75

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

Converted CpxTRS to CDT
0.00/0.75
0.00/0.75

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2) 0.00/0.75
add0(z0, Nil) → z0 0.00/0.75
notEmpty(Cons(z0, z1)) → True 0.00/0.75
notEmpty(Nil) → False 0.00/0.75
goal(z0, z1) → add0(z0, z1)
Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2)) 0.00/0.75
GOAL(z0, z1) → c4(ADD0(z0, z1))
S tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2)) 0.00/0.75
GOAL(z0, z1) → c4(ADD0(z0, z1))
K tuples:none
Defined Rule Symbols:

add0, notEmpty, goal

Defined Pair Symbols:

ADD0, GOAL

Compound Symbols:

c, c4

0.00/0.75
0.00/0.75

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c4(ADD0(z0, z1))
0.00/0.75
0.00/0.75

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2) 0.00/0.75
add0(z0, Nil) → z0 0.00/0.75
notEmpty(Cons(z0, z1)) → True 0.00/0.75
notEmpty(Nil) → False 0.00/0.75
goal(z0, z1) → add0(z0, z1)
Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
S tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
K tuples:none
Defined Rule Symbols:

add0, notEmpty, goal

Defined Pair Symbols:

ADD0

Compound Symbols:

c

0.00/0.75
0.00/0.75

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
We considered the (Usable) Rules:none
And the Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.75

POL(ADD0(x1, x2)) = x2    0.00/0.75
POL(Cons(x1, x2)) = [1] + x2    0.00/0.75
POL(Nil) = 0    0.00/0.75
POL(c(x1)) = x1   
0.00/0.75
0.00/0.75

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

add0(z0, Cons(z1, z2)) → add0(Cons(Cons(Nil, Nil), z0), z2) 0.00/0.75
add0(z0, Nil) → z0 0.00/0.75
notEmpty(Cons(z0, z1)) → True 0.00/0.75
notEmpty(Nil) → False 0.00/0.75
goal(z0, z1) → add0(z0, z1)
Tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
S tuples:none
K tuples:

ADD0(z0, Cons(z1, z2)) → c(ADD0(Cons(Cons(Nil, Nil), z0), z2))
Defined Rule Symbols:

add0, notEmpty, goal

Defined Pair Symbols:

ADD0

Compound Symbols:

c

0.00/0.75
0.00/0.75

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

The set S is empty
0.00/0.75
0.00/0.75

(8) BOUNDS(O(1), O(1))

0.00/0.75
0.00/0.75
0.00/0.78 EOF