YES(O(1), O(n^2)) 2.39/1.04 YES(O(1), O(n^2)) 2.39/1.05 2.39/1.05 2.39/1.05 2.39/1.05 2.39/1.06 2.39/1.06 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.39/1.06 2.39/1.06 2.39/1.06
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.39/1.06 
↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳6 CdtProblem
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↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳8 BOUNDS(O(1), O(1))
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(0) Obligation:

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

rev(nil) → nil 2.39/1.06
rev(.(x, y)) → ++(rev(y), .(x, nil)) 2.39/1.06
car(.(x, y)) → x 2.39/1.06
cdr(.(x, y)) → y 2.39/1.06
null(nil) → true 2.39/1.06
null(.(x, y)) → false 2.39/1.06
++(nil, y) → y 2.39/1.06
++(.(x, y), z) → .(x, ++(y, z))

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.39/1.06
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil)) 2.39/1.06
car(.(z0, z1)) → z0 2.39/1.06
cdr(.(z0, z1)) → z1 2.39/1.06
null(nil) → true 2.39/1.06
null(.(z0, z1)) → false 2.39/1.06
++(nil, z0) → z0 2.39/1.06
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
K tuples:none
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

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(3) 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.

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
We considered the (Usable) Rules:

rev(nil) → nil 2.39/1.06
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil)) 2.39/1.06
++(nil, z0) → z0 2.39/1.06
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
And the Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.39/1.06

POL(++(x1, x2)) = [5]    2.39/1.06
POL(++'(x1, x2)) = 0    2.39/1.06
POL(.(x1, x2)) = [1] + x1 + x2    2.39/1.06
POL(REV(x1)) = [2]x1    2.39/1.06
POL(c1(x1, x2)) = x1 + x2    2.39/1.06
POL(c7(x1)) = x1    2.39/1.06
POL(nil) = 0    2.39/1.06
POL(rev(x1)) = 0   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.39/1.06
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil)) 2.39/1.06
car(.(z0, z1)) → z0 2.39/1.06
cdr(.(z0, z1)) → z1 2.39/1.06
null(nil) → true 2.39/1.06
null(.(z0, z1)) → false 2.39/1.06
++(nil, z0) → z0 2.39/1.06
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:

++'(.(z0, z1), z2) → c7(++'(z1, z2))
K tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1))
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

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(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

++'(.(z0, z1), z2) → c7(++'(z1, z2))
We considered the (Usable) Rules:

rev(nil) → nil 2.39/1.06
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil)) 2.39/1.06
++(nil, z0) → z0 2.39/1.06
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
And the Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.39/1.06

POL(++(x1, x2)) = x1 + x2    2.39/1.06
POL(++'(x1, x2)) = [2]x1    2.39/1.06
POL(.(x1, x2)) = [1] + x2    2.39/1.06
POL(REV(x1)) = [2]x12    2.39/1.06
POL(c1(x1, x2)) = x1 + x2    2.39/1.06
POL(c7(x1)) = x1    2.39/1.06
POL(nil) = 0    2.39/1.06
POL(rev(x1)) = [2]x1   
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2.39/1.06

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.39/1.06
rev(.(z0, z1)) → ++(rev(z1), .(z0, nil)) 2.39/1.06
car(.(z0, z1)) → z0 2.39/1.06
cdr(.(z0, z1)) → z1 2.39/1.06
null(nil) → true 2.39/1.06
null(.(z0, z1)) → false 2.39/1.06
++(nil, z0) → z0 2.39/1.06
++(.(z0, z1), z2) → .(z0, ++(z1, z2))
Tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
S tuples:none
K tuples:

REV(.(z0, z1)) → c1(++'(rev(z1), .(z0, nil)), REV(z1)) 2.39/1.06
++'(.(z0, z1), z2) → c7(++'(z1, z2))
Defined Rule Symbols:

rev, car, cdr, null, ++

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c1, c7

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

The set S is empty
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(8) BOUNDS(O(1), O(1))

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2.78/1.24 EOF