(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^2).
The TRS R consists of the following rules:
half(0) → 0
half(s(s(x))) → s(half(x))
log(s(0)) → 0
log(s(s(x))) → s(log(s(half(x))))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
half(0) → 0
half(s(s(z0))) → s(half(z0))
log(s(0)) → 0
log(s(s(z0))) → s(log(s(half(z0))))
Tuples:
HALF(0) → c
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(0)) → c2
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:
HALF(0) → c
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(0)) → c2
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
K tuples:none
Defined Rule Symbols:
half, log
Defined Pair Symbols:
HALF, LOG
Compound Symbols:
c, c1, c2, c3
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
HALF(0) → c
LOG(s(0)) → c2
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
half(0) → 0
half(s(s(z0))) → s(half(z0))
log(s(0)) → 0
log(s(s(z0))) → s(log(s(half(z0))))
Tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
K tuples:none
Defined Rule Symbols:
half, log
Defined Pair Symbols:
HALF, LOG
Compound Symbols:
c1, c3
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
log(s(0)) → 0
log(s(s(z0))) → s(log(s(half(z0))))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
half(0) → 0
half(s(s(z0))) → s(half(z0))
Tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
K tuples:none
Defined Rule Symbols:
half
Defined Pair Symbols:
HALF, LOG
Compound Symbols:
c1, 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.
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
We considered the (Usable) Rules:
half(s(s(z0))) → s(half(z0))
half(0) → 0
And the Tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(HALF(x1)) = 0
POL(LOG(x1)) = x1
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(half(x1)) = x1
POL(s(x1)) = [1] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
half(0) → 0
half(s(s(z0))) → s(half(z0))
Tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:
HALF(s(s(z0))) → c1(HALF(z0))
K tuples:
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
Defined Rule Symbols:
half
Defined Pair Symbols:
HALF, LOG
Compound Symbols:
c1, c3
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
HALF(s(s(z0))) → c1(HALF(z0))
We considered the (Usable) Rules:
half(s(s(z0))) → s(half(z0))
half(0) → 0
And the Tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(HALF(x1)) = [2]x1
POL(LOG(x1)) = x12
POL(c1(x1)) = x1
POL(c3(x1, x2)) = x1 + x2
POL(half(x1)) = x1
POL(s(x1)) = [2] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
half(0) → 0
half(s(s(z0))) → s(half(z0))
Tuples:
HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
S tuples:none
K tuples:
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
HALF(s(s(z0))) → c1(HALF(z0))
Defined Rule Symbols:
half
Defined Pair Symbols:
HALF, LOG
Compound Symbols:
c1, c3
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)