(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__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:
A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C)
A__F(z0, z1, z2) → c2
A__C → c3
A__C → c4
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
MARK(c) → c6(A__C)
MARK(b) → c7
S tuples:
A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C)
A__F(z0, z1, z2) → c2
A__C → c3
A__C → c4
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
MARK(c) → c6(A__C)
MARK(b) → c7
K tuples:none
Defined Rule Symbols:
a__f, a__c, mark
Defined Pair Symbols:
A__F, A__C, MARK
Compound Symbols:
c1, c2, c3, c4, c5, c6, c7
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 6 trailing nodes:
A__F(b, z0, c) → c1(A__F(z0, a__c, z0), A__C)
A__C → c4
MARK(c) → c6(A__C)
A__F(z0, z1, z2) → c2
MARK(b) → c7
A__C → c3
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
S tuples:
MARK(f(z0, z1, z2)) → c5(A__F(z0, mark(z1), z2), MARK(z1))
K tuples:none
Defined Rule Symbols:
a__f, a__c, mark
Defined Pair Symbols:
MARK
Compound Symbols:
c5
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
Tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
S tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
K tuples:none
Defined Rule Symbols:
a__f, a__c, mark
Defined Pair Symbols:
MARK
Compound Symbols:
c5
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
a__f(b, z0, c) → a__f(z0, a__c, z0)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__c → b
a__c → c
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(c) → a__c
mark(b) → b
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
S tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
MARK
Compound Symbols:
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.
MARK(f(z0, z1, z2)) → c5(MARK(z1))
We considered the (Usable) Rules:none
And the Tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(MARK(x1)) = x1
POL(c5(x1)) = x1
POL(f(x1, x2, x3)) = [1] + x2
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
S tuples:none
K tuples:
MARK(f(z0, z1, z2)) → c5(MARK(z1))
Defined Rule Symbols:none
Defined Pair Symbols:
MARK
Compound Symbols:
c5
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)