(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:
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))
double(x) → +(x, x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
double(0) → 0
double(s(z0)) → s(s(double(z0)))
double(z0) → +(z0, z0)
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))
Tuples:
DOUBLE(0) → c
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, 0) → c3
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:
DOUBLE(0) → c
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, 0) → c3
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:
double, +
Defined Pair Symbols:
DOUBLE, +'
Compound Symbols:
c, c1, c2, c3, c4, c5
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
+'(z0, 0) → c3
DOUBLE(0) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
double(0) → 0
double(s(z0)) → s(s(double(z0)))
double(z0) → +(z0, z0)
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))
Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:
double, +
Defined Pair Symbols:
DOUBLE, +'
Compound Symbols:
c1, c2, c4, c5
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
double(0) → 0
double(s(z0)) → s(s(double(z0)))
double(z0) → +(z0, z0)
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))
+(s(z0), z1) → s(+(z0, z1))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
DOUBLE, +'
Compound Symbols:
c1, c2, c4, c5
(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.
DOUBLE(z0) → c2(+'(z0, z0))
We considered the (Usable) Rules:none
And the Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(+'(x1, x2)) = 0
POL(DOUBLE(x1)) = [1]
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = 0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:
DOUBLE(z0) → c2(+'(z0, z0))
Defined Rule Symbols:none
Defined Pair Symbols:
DOUBLE, +'
Compound Symbols:
c1, c2, c4, c5
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DOUBLE(s(z0)) → c1(DOUBLE(z0))
We considered the (Usable) Rules:none
And the Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(+'(x1, x2)) = [1]
POL(DOUBLE(x1)) = [1] + x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:
DOUBLE(z0) → c2(+'(z0, z0))
DOUBLE(s(z0)) → c1(DOUBLE(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
DOUBLE, +'
Compound Symbols:
c1, c2, c4, c5
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(+'(x1, x2)) = [2] + [2]x1 + x2
POL(DOUBLE(x1)) = [2] + [3]x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(s(x1)) = [2] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:none
K tuples:
DOUBLE(z0) → c2(+'(z0, z0))
DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))
Defined Rule Symbols:none
Defined Pair Symbols:
DOUBLE, +'
Compound Symbols:
c1, c2, c4, c5
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)