(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:
a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c
Rewrite Strategy: INNERMOST
(1) CpxTrsMatchBoundsProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 3.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[a__g_1|0, a__c|0, a__h_1|0, mark_1|0, a__h_1|1, g_1|1, d|1, c|1, h_1|1, a__g_1|1, a__c|1, a__h_1|2, g_1|2, d|2, c|2, h_1|2, h_1|3]
1→3[a__g_1|1, a__h_1|2, g_1|2, h_1|3]
2→2[d|0, c|0, g_1|0, h_1|0]
3→2[c|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:
a__g(z0) → a__h(z0)
a__g(z0) → g(z0)
a__c → d
a__c → c
a__h(d) → a__g(c)
a__h(z0) → h(z0)
mark(g(z0)) → a__g(z0)
mark(h(z0)) → a__h(z0)
mark(c) → a__c
mark(d) → d
Tuples:
A__G(z0) → c1(A__H(z0))
A__G(z0) → c2
A__C → c3
A__C → c4
A__H(d) → c5(A__G(c))
A__H(z0) → c6
MARK(g(z0)) → c7(A__G(z0))
MARK(h(z0)) → c8(A__H(z0))
MARK(c) → c9(A__C)
MARK(d) → c10
S tuples:
A__G(z0) → c1(A__H(z0))
A__G(z0) → c2
A__C → c3
A__C → c4
A__H(d) → c5(A__G(c))
A__H(z0) → c6
MARK(g(z0)) → c7(A__G(z0))
MARK(h(z0)) → c8(A__H(z0))
MARK(c) → c9(A__C)
MARK(d) → c10
K tuples:none
Defined Rule Symbols:
a__g, a__c, a__h, mark
Defined Pair Symbols:
A__G, A__C, A__H, MARK
Compound Symbols:
c1, c2, c3, c4, c5, c6, c7, c8, c9, c10
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 2 leading nodes:
MARK(h(z0)) → c8(A__H(z0))
MARK(g(z0)) → c7(A__G(z0))
Removed 6 trailing nodes:
MARK(c) → c9(A__C)
A__H(z0) → c6
A__G(z0) → c2
A__C → c4
MARK(d) → c10
A__C → c3
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__g(z0) → a__h(z0)
a__g(z0) → g(z0)
a__c → d
a__c → c
a__h(d) → a__g(c)
a__h(z0) → h(z0)
mark(g(z0)) → a__g(z0)
mark(h(z0)) → a__h(z0)
mark(c) → a__c
mark(d) → d
Tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
S tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
K tuples:none
Defined Rule Symbols:
a__g, a__c, a__h, mark
Defined Pair Symbols:
A__G, A__H
Compound Symbols:
c1, c5
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
a__g(z0) → a__h(z0)
a__g(z0) → g(z0)
a__c → d
a__c → c
a__h(d) → a__g(c)
a__h(z0) → h(z0)
mark(g(z0)) → a__g(z0)
mark(h(z0)) → a__h(z0)
mark(c) → a__c
mark(d) → d
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
S tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
A__G, A__H
Compound Symbols:
c1, c5
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^3)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
We considered the (Usable) Rules:none
And the Tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(A__G(x1)) = [1] + x1 + x12 + x13
POL(A__H(x1)) = x1 + x12 + x13
POL(c) = 0
POL(c1(x1)) = x1
POL(c5(x1)) = x1
POL(d) = [1]
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
S tuples:none
K tuples:
A__G(z0) → c1(A__H(z0))
A__H(d) → c5(A__G(c))
Defined Rule Symbols:none
Defined Pair Symbols:
A__G, A__H
Compound Symbols:
c1, c5
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)