### (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, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

Rewrite Strategy: FULL

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

The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
f(f(x, y), z) → f(x, f(y, z))

### (2) 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(g(x, y), z) → g(f(x, z), f(y, z))
f(2, g(x, y)) → y
f(1, g(x, y)) → x
f(i(x), y) → i(x)
f(0, y) → y
f(x, 0) → x

Rewrite Strategy: FULL

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

Converted rc-obligation to irc-obligation.

As the TRS does not nest defined symbols, we have rc = irc.

### (4) 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(g(x, y), z) → g(f(x, z), f(y, z))
f(2, g(x, y)) → y
f(1, g(x, y)) → x
f(i(x), y) → i(x)
f(0, y) → y
f(x, 0) → x

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(g(z0, z1), z2) → g(f(z0, z2), f(z1, z2))
f(2, g(z0, z1)) → z1
f(1, g(z0, z1)) → z0
f(i(z0), z1) → i(z0)
f(0, z0) → z0
f(z0, 0) → z0
Tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
F(2, g(z0, z1)) → c1
F(1, g(z0, z1)) → c2
F(i(z0), z1) → c3
F(0, z0) → c4
F(z0, 0) → c5
S tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
F(2, g(z0, z1)) → c1
F(1, g(z0, z1)) → c2
F(i(z0), z1) → c3
F(0, z0) → c4
F(z0, 0) → c5
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 5 trailing nodes:

F(i(z0), z1) → c3
F(1, g(z0, z1)) → c2
F(z0, 0) → c5
F(0, z0) → c4
F(2, g(z0, z1)) → c1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(g(z0, z1), z2) → g(f(z0, z2), f(z1, z2))
f(2, g(z0, z1)) → z1
f(1, g(z0, z1)) → z0
f(i(z0), z1) → i(z0)
f(0, z0) → z0
f(z0, 0) → z0
Tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
S tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(g(z0, z1), z2) → g(f(z0, z2), f(z1, z2))
f(2, g(z0, z1)) → z1
f(1, g(z0, z1)) → z0
f(i(z0), z1) → i(z0)
f(0, z0) → z0
f(z0, 0) → z0

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
S tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

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.

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

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

POL(F(x1, x2)) = [2] + [2]x1
POL(c(x1, x2)) = x1 + x2
POL(g(x1, x2)) = [2] + x1 + x2

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
S tuples:none
K tuples:

F(g(z0, z1), z2) → c(F(z0, z2), F(z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

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

The set S is empty