### (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(x, g(x)) → x
f(x, h(y)) → f(h(x), y)

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(x, g(x)) → x
f(x, h(y)) → f(h(x), y)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, g(z0)) → z0
f(z0, h(z1)) → f(h(z0), z1)
Tuples:

F(z0, g(z0)) → c
F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:

F(z0, g(z0)) → c
F(z0, h(z1)) → c1(F(h(z0), z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(z0, g(z0)) → c

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, g(z0)) → z0
f(z0, h(z1)) → f(h(z0), z1)
Tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0, g(z0)) → z0
f(z0, h(z1)) → f(h(z0), z1)

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

### (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.

F(z0, h(z1)) → c1(F(h(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = x2
POL(c1(x1)) = x1
POL(h(x1)) = [1] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
S tuples:none
K tuples:

F(z0, h(z1)) → c1(F(h(z0), z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty