### (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:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, 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:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
S tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
S tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c, c1

### (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.

F(z0, z1, s(z2)) → c1(F(0, 1, z2))
We considered the (Usable) Rules:none
And the Tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(1) = 0
POL(F(x1, x2, x3)) = [2]x3
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(s(x1)) = [2] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
S tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
K tuples:

F(z0, z1, s(z2)) → c1(F(0, 1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c, c1

### (9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
Now S is empty