(0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

a__f(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__bc
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__bb

Rewrite Strategy: FULL

(1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
a__f(X, g(X), [])

The defined contexts are:
a__g([])

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

(2) 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(X, g(X), Y) → a__f(Y, Y, Y)
a__g(b) → c
a__bc
mark(f(X1, X2, X3)) → a__f(X1, X2, X3)
mark(g(X)) → a__g(mark(X))
mark(b) → a__b
mark(c) → c
a__f(X1, X2, X3) → f(X1, X2, X3)
a__g(X) → g(X)
a__bb

Rewrite Strategy: INNERMOST

(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__bc
a__bb
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

A__F(z0, g(z0), z1) → c1(A__F(z1, z1, z1))
A__F(z0, z1, z2) → c2
A__G(b) → c3
A__G(z0) → c4
A__Bc5
A__Bc6
MARK(f(z0, z1, z2)) → c7(A__F(z0, z1, z2))
MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
MARK(b) → c9(A__B)
MARK(c) → c10
S tuples:

A__F(z0, g(z0), z1) → c1(A__F(z1, z1, z1))
A__F(z0, z1, z2) → c2
A__G(b) → c3
A__G(z0) → c4
A__Bc5
A__Bc6
MARK(f(z0, z1, z2)) → c7(A__F(z0, z1, z2))
MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
MARK(b) → c9(A__B)
MARK(c) → c10
K tuples:none
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

A__F, A__G, A__B, MARK

Compound Symbols:

c1, c2, c3, c4, c5, c6, c7, c8, c9, c10

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 9 trailing nodes:

MARK(f(z0, z1, z2)) → c7(A__F(z0, z1, z2))
A__F(z0, g(z0), z1) → c1(A__F(z1, z1, z1))
A__Bc5
A__G(z0) → c4
A__G(b) → c3
A__F(z0, z1, z2) → c2
A__Bc6
MARK(b) → c9(A__B)
MARK(c) → c10

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__bc
a__bb
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
S tuples:

MARK(g(z0)) → c8(A__G(mark(z0)), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c8

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__bc
a__bb
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c
Tuples:

MARK(g(z0)) → c8(MARK(z0))
S tuples:

MARK(g(z0)) → c8(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__f, a__g, a__b, mark

Defined Pair Symbols:

MARK

Compound Symbols:

c8

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

a__f(z0, g(z0), z1) → a__f(z1, z1, z1)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__g(b) → c
a__g(z0) → g(z0)
a__bc
a__bb
mark(f(z0, z1, z2)) → a__f(z0, z1, z2)
mark(g(z0)) → a__g(mark(z0))
mark(b) → a__b
mark(c) → c

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(g(z0)) → c8(MARK(z0))
S tuples:

MARK(g(z0)) → c8(MARK(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c8

(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.

MARK(g(z0)) → c8(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

MARK(g(z0)) → c8(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MARK(x1)) = x1   
POL(c8(x1)) = x1   
POL(g(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MARK(g(z0)) → c8(MARK(z0))
S tuples:none
K tuples:

MARK(g(z0)) → c8(MARK(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MARK

Compound Symbols:

c8

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(14) BOUNDS(1, 1)