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

(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(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
a__b → b

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(a, z0, z0) → a__f(z0, a__b, b)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__b → a
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(b) → a__b
mark(a) → a

Tuples:

A__F(a, z0, z0) → c(A__F(z0, a__b, b), A__B)
A__F(z0, z1, z2) → c1
A__B → c2
A__B → c3
MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
MARK(b) → c5(A__B)
MARK(a) → c6

S tuples:

A__F(a, z0, z0) → c(A__F(z0, a__b, b), A__B)
A__F(z0, z1, z2) → c1
A__B → c2
A__B → c3
MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
MARK(b) → c5(A__B)
MARK(a) → c6

K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

A__F, A__B, MARK

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

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

Removed 5 trailing nodes:

MARK(b) → c5(A__B)
MARK(a) → c6
A__B → c3
A__F(z0, z1, z2) → c1
A__B → c2

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(a, z0, z0) → a__f(z0, a__b, b)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__b → a
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(b) → a__b
mark(a) → a

Tuples:

A__F(a, z0, z0) → c(A__F(z0, a__b, b), A__B)
MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))

S tuples:

A__F(a, z0, z0) → c(A__F(z0, a__b, b), A__B)
MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))

K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

A__F, MARK

Compound Symbols:

c, c4

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(a, z0, z0) → a__f(z0, a__b, b)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__b → a
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(b) → a__b
mark(a) → a

Tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

S tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

K tuples:none
Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c

(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.

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))

We considered the (Usable) Rules:none
And the Tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__F(x₁, x₂, x₃)) = x₃   
POL(MARK(x₁)) = x₁   
POL(a) = 0   
POL(a__b) = [1]   
POL(a__f(x₁, x₂, x₃)) = 0   
POL(b) = 0   
POL(c(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂, x₃)) = [1] + x₁ + x₂ + x₃   
POL(mark(x₁)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(a, z0, z0) → a__f(z0, a__b, b)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__b → a
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(b) → a__b
mark(a) → a

Tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

S tuples:

A__F(a, z0, z0) → c(A__F(z0, a__b, b))

K tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))

Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c

(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.

A__F(a, z0, z0) → c(A__F(z0, a__b, b))

We considered the (Usable) Rules:none
And the Tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(A__F(x₁, x₂, x₃)) = x₁ + x₃   
POL(MARK(x₁)) = [2]x₁   
POL(a) = [2]   
POL(a__b) = [2]   
POL(a__f(x₁, x₂, x₃)) = 0   
POL(b) = [1]   
POL(c(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(mark(x₁)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__f(a, z0, z0) → a__f(z0, a__b, b)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__b → a
a__b → b
mark(f(z0, z1, z2)) → a__f(z0, mark(z1), z2)
mark(b) → a__b
mark(a) → a

Tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

S tuples:none
K tuples:

MARK(f(z0, z1, z2)) → c4(A__F(z0, mark(z1), z2), MARK(z1))
A__F(a, z0, z0) → c(A__F(z0, a__b, b))

Defined Rule Symbols:

a__f, a__b, mark

Defined Pair Symbols:

MARK, A__F

Compound Symbols:

c4, c

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

The set S is empty

(12) BOUNDS(1, 1)