(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:
c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → y
Rewrite Strategy: INNERMOST
(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 2.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1]
transitions:
a0(0, 0) → 0
00() → 0
c0(0) → 1
01() → 5
c1(5) → 4
c1(4) → 3
c1(3) → 2
a1(2, 0) → 1
02() → 8
a2(5, 8) → 7
c2(7) → 6
c2(6) → 2
0 → 1
5 → 4
4 → 3
3 → 2
6 → 2
7 → 6
(2) BOUNDS(1, n^1)
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0
Tuples:
C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
C(z0) → c3
S tuples:
C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
C(z0) → c3
K tuples:none
Defined Rule Symbols:
c
Defined Pair Symbols:
C
Compound Symbols:
c1, c2, c3
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
C(c(c(z0))) → c1(C(c(a(z0, 0))), C(a(z0, 0)))
Removed 1 trailing nodes:
C(z0) → c3
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0
Tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
S tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)), C(0))
K tuples:none
Defined Rule Symbols:
c
Defined Pair Symbols:
C
Compound Symbols:
c2
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(c(z0))) → c(c(a(z0, 0)))
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0
Tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
S tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
K tuples:none
Defined Rule Symbols:
c
Defined Pair Symbols:
C
Compound Symbols:
c2
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
c(c(c(z0))) → c(c(a(z0, 0)))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0
Tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
S tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
K tuples:none
Defined Rule Symbols:
c
Defined Pair Symbols:
C
Compound Symbols:
c2
(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.
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
We considered the (Usable) Rules:
c(z0) → z0
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
And the Tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(C(x1)) = [1] + x1
POL(a(x1, x2)) = [1] + x1
POL(c(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(a(a(0, z0), z1)) → a(c(c(c(0))), z1)
c(z0) → z0
Tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
S tuples:none
K tuples:
C(a(a(0, z0), z1)) → c2(C(c(c(0))), C(c(0)))
Defined Rule Symbols:
c
Defined Pair Symbols:
C
Compound Symbols:
c2
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)