(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:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)
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 4.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1]
transitions:
a0() → 0
b0() → 0
s0(0) → 0
c0() → 0
f0(0, 0) → 1
a1() → 2
b1() → 3
f1(2, 3) → 1
a1() → 5
s1(5) → 4
c1() → 6
f1(4, 6) → 1
f1(0, 6) → 1
a1() → 7
f1(2, 7) → 1
a2() → 8
b2() → 9
f2(8, 9) → 1
a2() → 11
s2(11) → 10
c2() → 12
f2(10, 12) → 1
f2(5, 12) → 1
a3() → 14
s3(14) → 13
c3() → 15
f3(13, 15) → 1
f3(11, 15) → 1
c4() → 16
f4(14, 16) → 1
(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:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(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.
F(c, c) → c4(F(a, a))
We considered the (Usable) Rules:none
And the Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x1
POL(a) = 0
POL(b) = 0
POL(c) = [1]
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(s(x1)) = x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
K tuples:
F(c, c) → c4(F(a, a))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(9) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(a, b) → c2(F(s(a), c))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(s(z0), c) → c3(F(z0, c))
K tuples:
F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(11) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use forward instantiation to replace
F(
s(
z0),
c) →
c3(
F(
z0,
c)) by
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(c, c) → c4(F(a, a))
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))
S tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))
K tuples:
F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c4, c3
(13) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing nodes:
F(a, b) → c2(F(s(a), c))
F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(s(c), c) → c3(F(c, c))
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
S tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c3
(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
F(s(s(y0)), c) → c3(F(s(y0), c))
We considered the (Usable) Rules:none
And the Tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x1
POL(c) = 0
POL(c3(x1)) = x1
POL(s(x1)) = [1] + x1
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
S tuples:none
K tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c3
(17) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(18) BOUNDS(1, 1)