YES(O(1), O(n^1)) 3.97/1.43 YES(O(1), O(n^1)) 3.97/1.48 3.97/1.48 3.97/1.48 3.97/1.48 3.97/1.48 3.97/1.48 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 3.97/1.48 3.97/1.48 3.97/1.48
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtUnreachableProof (⇔)
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↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳8 CdtProblem
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↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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3.97/1.48

(0) Obligation:

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

circ(cons(a, s), t) → cons(msubst(a, t), circ(s, t)) 3.97/1.48
circ(cons(lift, s), cons(a, t)) → cons(a, circ(s, t)) 3.97/1.48
circ(cons(lift, s), cons(lift, t)) → cons(lift, circ(s, t)) 3.97/1.48
circ(circ(s, t), u) → circ(s, circ(t, u)) 3.97/1.48
circ(s, id) → s 3.97/1.48
circ(id, s) → s 3.97/1.48
circ(cons(lift, s), circ(cons(lift, t), u)) → circ(cons(lift, circ(s, t)), u) 3.97/1.48
subst(a, id) → a 3.97/1.48
msubst(a, id) → a 3.97/1.48
msubst(msubst(a, s), t) → msubst(a, circ(s, t))

Rewrite Strategy: INNERMOST
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3.97/1.48

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

Converted CpxTRS to CDT
3.97/1.48
3.97/1.48

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2)) 3.97/1.48
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2)) 3.97/1.48
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1)) 3.97/1.48
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2)) 3.97/1.48
circ(z0, id) → z0 3.97/1.48
circ(id, z0) → z0 3.97/1.48
circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2) 3.97/1.48
subst(z0, id) → z0 3.97/1.48
msubst(z0, id) → z0 3.97/1.48
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 3.97/1.48
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 3.97/1.48
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 3.97/1.48
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 3.97/1.48
CIRC(cons(lift, z0), circ(cons(lift, z1), z2)) → c6(CIRC(cons(lift, circ(z0, z1)), z2), CIRC(z0, z1)) 3.97/1.48
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 3.97/1.48
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 3.97/1.48
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 3.97/1.48
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 3.97/1.48
CIRC(cons(lift, z0), circ(cons(lift, z1), z2)) → c6(CIRC(cons(lift, circ(z0, z1)), z2), CIRC(z0, z1)) 3.97/1.48
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
K tuples:none
Defined Rule Symbols:

circ, subst, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c6, c9

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3.97/1.48

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

CIRC(cons(lift, z0), circ(cons(lift, z1), z2)) → c6(CIRC(cons(lift, circ(z0, z1)), z2), CIRC(z0, z1))
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4.34/1.52

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2)) 4.34/1.52
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2)) 4.34/1.52
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1)) 4.34/1.52
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2)) 4.34/1.52
circ(z0, id) → z0 4.34/1.52
circ(id, z0) → z0 4.34/1.52
circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2) 4.34/1.52
subst(z0, id) → z0 4.34/1.52
msubst(z0, id) → z0 4.34/1.52
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
K tuples:none
Defined Rule Symbols:

circ, subst, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

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4.34/1.52

(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.

CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
We considered the (Usable) Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2)) 4.34/1.52
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2)) 4.34/1.52
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1)) 4.34/1.52
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2)) 4.34/1.52
circ(z0, id) → z0 4.34/1.52
circ(id, z0) → z0 4.34/1.52
msubst(z0, id) → z0 4.34/1.52
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
And the Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 4.34/1.52

POL(CIRC(x1, x2)) = [4]x1    4.34/1.52
POL(MSUBST(x1, x2)) = [4]x1    4.34/1.52
POL(c(x1, x2)) = x1 + x2    4.34/1.52
POL(c1(x1)) = x1    4.34/1.52
POL(c2(x1)) = x1    4.34/1.52
POL(c3(x1, x2)) = x1 + x2    4.34/1.52
POL(c9(x1, x2)) = x1 + x2    4.34/1.52
POL(circ(x1, x2)) = [1] + [4]x1 + x2    4.34/1.52
POL(cons(x1, x2)) = x1 + x2    4.34/1.52
POL(id) = 0    4.34/1.52
POL(lift) = 0    4.34/1.52
POL(msubst(x1, x2)) = x1 + [2]x2   
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4.34/1.52

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2)) 4.34/1.52
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2)) 4.34/1.52
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1)) 4.34/1.52
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2)) 4.34/1.52
circ(z0, id) → z0 4.34/1.52
circ(id, z0) → z0 4.34/1.52
circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2) 4.34/1.52
subst(z0, id) → z0 4.34/1.52
msubst(z0, id) → z0 4.34/1.52
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
K tuples:

CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2))
Defined Rule Symbols:

circ, subst, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

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4.34/1.52

(7) 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.

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
We considered the (Usable) Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2)) 4.34/1.52
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2)) 4.34/1.52
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1)) 4.34/1.52
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2)) 4.34/1.52
circ(z0, id) → z0 4.34/1.52
circ(id, z0) → z0 4.34/1.52
msubst(z0, id) → z0 4.34/1.52
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
And the Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 4.34/1.52

POL(CIRC(x1, x2)) = [2]x1    4.34/1.52
POL(MSUBST(x1, x2)) = [5] + x1    4.34/1.52
POL(c(x1, x2)) = x1 + x2    4.34/1.52
POL(c1(x1)) = x1    4.34/1.52
POL(c2(x1)) = x1    4.34/1.52
POL(c3(x1, x2)) = x1 + x2    4.34/1.52
POL(c9(x1, x2)) = x1 + x2    4.34/1.52
POL(circ(x1, x2)) = [1] + [2]x1 + x2    4.34/1.52
POL(cons(x1, x2)) = [4] + x1 + x2    4.34/1.52
POL(id) = 0    4.34/1.52
POL(lift) = 0    4.34/1.52
POL(msubst(x1, x2)) = [2] + [5]x1 + [4]x2   
4.34/1.52
4.34/1.52

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

circ(cons(z0, z1), z2) → cons(msubst(z0, z2), circ(z1, z2)) 4.34/1.52
circ(cons(lift, z0), cons(z1, z2)) → cons(z1, circ(z0, z2)) 4.34/1.52
circ(cons(lift, z0), cons(lift, z1)) → cons(lift, circ(z0, z1)) 4.34/1.52
circ(circ(z0, z1), z2) → circ(z0, circ(z1, z2)) 4.34/1.52
circ(z0, id) → z0 4.34/1.52
circ(id, z0) → z0 4.34/1.52
circ(cons(lift, z0), circ(cons(lift, z1), z2)) → circ(cons(lift, circ(z0, z1)), z2) 4.34/1.52
subst(z0, id) → z0 4.34/1.52
msubst(z0, id) → z0 4.34/1.52
msubst(msubst(z0, z1), z2) → msubst(z0, circ(z1, z2))
Tuples:

CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
S tuples:none
K tuples:

CIRC(circ(z0, z1), z2) → c3(CIRC(z0, circ(z1, z2)), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(z0, z1), z2) → c(MSUBST(z0, z2), CIRC(z1, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(z1, z2)) → c1(CIRC(z0, z2)) 4.34/1.52
CIRC(cons(lift, z0), cons(lift, z1)) → c2(CIRC(z0, z1)) 4.34/1.52
MSUBST(msubst(z0, z1), z2) → c9(MSUBST(z0, circ(z1, z2)), CIRC(z1, z2))
Defined Rule Symbols:

circ, subst, msubst

Defined Pair Symbols:

CIRC, MSUBST

Compound Symbols:

c, c1, c2, c3, c9

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4.34/1.52

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

The set S is empty
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4.34/1.52

(10) BOUNDS(O(1), O(1))

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4.66/1.62 EOF