(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)
Tuples:
REV(a) → c
REV(b) → c1
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:
REV(a) → c
REV(b) → c1
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:
rev
Defined Pair Symbols:
REV
Compound Symbols:
c, c1, c2, c3
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
REV(b) → c1
REV(a) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)
Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:
rev
Defined Pair Symbols:
REV
Compound Symbols:
c2, c3
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
rev(a) → a
rev(b) → b
rev(++(z0, z1)) → ++(rev(z1), rev(z0))
rev(++(z0, z0)) → rev(z0)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
REV
Compound Symbols:
c2, c3
(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.
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
We considered the (Usable) Rules:none
And the Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(++(x1, x2)) = [1] + x1 + x2
POL(REV(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
S tuples:none
K tuples:
REV(++(z0, z1)) → c2(REV(z1), REV(z0))
REV(++(z0, z0)) → c3(REV(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
REV
Compound Symbols:
c2, c3
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)