YES(O(1), O(n^2)) 2.51/1.06 YES(O(1), O(n^2)) 2.51/1.10 2.51/1.10 2.51/1.10 2.51/1.10 2.51/1.10 2.51/1.10 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.51/1.10 2.51/1.10 2.51/1.10
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2.51/1.10

(0) Obligation:

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

rev(nil) → nil 2.51/1.10
rev(rev(x)) → x 2.51/1.10
rev(++(x, y)) → ++(rev(y), rev(x)) 2.51/1.10
++(nil, y) → y 2.51/1.10
++(x, nil) → x 2.51/1.10
++(.(x, y), z) → .(x, ++(y, z)) 2.51/1.10
++(x, ++(y, z)) → ++(++(x, y), z) 2.51/1.10
make(x) → .(x, nil)

Rewrite Strategy: INNERMOST
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2.51/1.10

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

Converted CpxTRS to CDT
2.51/1.10
2.51/1.10

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.51/1.10
rev(rev(z0)) → z0 2.51/1.10
rev(++(z0, z1)) → ++(rev(z1), rev(z0)) 2.51/1.10
++(nil, z0) → z0 2.51/1.10
++(z0, nil) → z0 2.51/1.10
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 2.51/1.10
++(z0, ++(z1, z2)) → ++(++(z0, z1), z2) 2.51/1.10
make(z0) → .(z0, nil)
Tuples:

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0)) 2.51/1.10
++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.51/1.10
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0)) 2.51/1.10
++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.51/1.10
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
K tuples:none
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

REV, ++'

Compound Symbols:

c2, c5, c6

2.51/1.10
2.51/1.10

(3) CdtUnreachableProof (EQUIVALENT transformation)

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

REV(++(z0, z1)) → c2(++'(rev(z1), rev(z0)), REV(z1), REV(z0))
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2.51/1.10

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.51/1.10
rev(rev(z0)) → z0 2.51/1.10
rev(++(z0, z1)) → ++(rev(z1), rev(z0)) 2.51/1.10
++(nil, z0) → z0 2.51/1.10
++(z0, nil) → z0 2.51/1.10
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 2.51/1.10
++(z0, ++(z1, z2)) → ++(++(z0, z1), z2) 2.51/1.10
make(z0) → .(z0, nil)
Tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.51/1.10
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.51/1.10
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
K tuples:none
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

++'

Compound Symbols:

c5, c6

2.51/1.10
2.51/1.10

(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.

++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
We considered the (Usable) Rules:

++(nil, z0) → z0 2.51/1.10
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 2.51/1.10
++(z0, ++(z1, z2)) → ++(++(z0, z1), z2) 2.51/1.10
++(z0, nil) → z0
And the Tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.51/1.10
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.88/1.11

POL(++(x1, x2)) = [4] + x1 + [4]x2    2.88/1.11
POL(++'(x1, x2)) = [5] + [2]x2    2.88/1.11
POL(.(x1, x2)) = x1    2.88/1.11
POL(c5(x1)) = x1    2.88/1.11
POL(c6(x1, x2)) = x1 + x2    2.88/1.11
POL(nil) = 0   
2.88/1.11
2.88/1.11

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.88/1.11
rev(rev(z0)) → z0 2.88/1.11
rev(++(z0, z1)) → ++(rev(z1), rev(z0)) 2.88/1.11
++(nil, z0) → z0 2.88/1.11
++(z0, nil) → z0 2.88/1.11
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 2.88/1.11
++(z0, ++(z1, z2)) → ++(++(z0, z1), z2) 2.88/1.11
make(z0) → .(z0, nil)
Tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.88/1.11
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2))
K tuples:

++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

++'

Compound Symbols:

c5, c6

2.88/1.11
2.88/1.11

(7) 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) → c5(++'(z1, z2))
We considered the (Usable) Rules:

++(nil, z0) → z0 2.88/1.11
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 2.88/1.11
++(z0, ++(z1, z2)) → ++(++(z0, z1), z2) 2.88/1.11
++(z0, nil) → z0
And the Tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.88/1.11
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.88/1.11

POL(++(x1, x2)) = [2] + x1 + x2    2.88/1.11
POL(++'(x1, x2)) = x1 + [3]x22 + x1·x2    2.88/1.11
POL(.(x1, x2)) = [2] + x1 + x2    2.88/1.11
POL(c5(x1)) = x1    2.88/1.11
POL(c6(x1, x2)) = x1 + x2    2.88/1.11
POL(nil) = [3]   
2.88/1.11
2.88/1.11

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

rev(nil) → nil 2.88/1.11
rev(rev(z0)) → z0 2.88/1.11
rev(++(z0, z1)) → ++(rev(z1), rev(z0)) 2.88/1.11
++(nil, z0) → z0 2.88/1.11
++(z0, nil) → z0 2.88/1.11
++(.(z0, z1), z2) → .(z0, ++(z1, z2)) 2.88/1.11
++(z0, ++(z1, z2)) → ++(++(z0, z1), z2) 2.88/1.11
make(z0) → .(z0, nil)
Tuples:

++'(.(z0, z1), z2) → c5(++'(z1, z2)) 2.88/1.11
++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1))
S tuples:none
K tuples:

++'(z0, ++(z1, z2)) → c6(++'(++(z0, z1), z2), ++'(z0, z1)) 2.88/1.11
++'(.(z0, z1), z2) → c5(++'(z1, z2))
Defined Rule Symbols:

rev, ++, make

Defined Pair Symbols:

++'

Compound Symbols:

c5, c6

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2.88/1.11

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

The set S is empty
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2.88/1.11

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

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2.88/1.11
3.21/1.24 EOF