### (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:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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:
:(:(x, y), z) → :(x, :(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:

:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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:

:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
:'(z0, +(z1, f(z2))) → c1(:'(g(z0, z2), +(z1, a)))
S tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
:'(z0, +(z1, f(z2))) → c1(:'(g(z0, z2), +(z1, a)))
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

:'(z0, +(z1, f(z2))) → c1(:'(g(z0, z2), +(z1, a)))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))
Tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
S tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
K tuples:none
Defined Rule Symbols:

:

Defined Pair Symbols:

:'

Compound Symbols:

c

### (9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

:(+(z0, z1), z2) → +(:(z0, z2), :(z1, z2))
:(z0, +(z1, f(z2))) → :(g(z0, z2), +(z1, a))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
S tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

:'

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.

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

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

POL(+(x1, x2)) = [2] + x1 + x2
POL(:'(x1, x2)) = [2] + [2]x1
POL(c(x1, x2)) = x1 + x2

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
S tuples:none
K tuples:

:'(+(z0, z1), z2) → c(:'(z0, z2), :'(z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

:'

Compound Symbols:

c

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

The set S is empty