### (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(a, X, X) → a__f(X, a__b, b)
a__ba
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__bb

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS is a non-duplicating overlay system, 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(a, X, X) → a__f(X, a__b, b)
a__ba
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__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(a, z0, z0) → a__f(z0, a__b, b)
a__f(z0, z1, z2) → f(z0, z1, z2)
a__ba
a__bb
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__Bc2
A__Bc3
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__Bc2
A__Bc3
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

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

Removed 5 trailing nodes:

MARK(a) → c6
A__Bc3
A__Bc2
A__F(z0, z1, z2) → c1
MARK(b) → c5(A__B)

### (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__ba
a__bb
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

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

Removed 1 trailing tuple parts

### (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__ba
a__bb
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

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

### (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__ba
a__bb
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

### (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.

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)) = x1
POL(a) = [1]
POL(a__b) = 0
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)) = x1 + x2 + x3
POL(mark(x1)) = 0

### (12) 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__ba
a__bb
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

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

The set S is empty