### (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))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

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))
+(f(x), +(f(y), z)) → +(f(+(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:

+(f(x), f(y)) → f(+(x, y))

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(x), f(y)) → f(+(x, y))

Rewrite Strategy: INNERMOST

### (5) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : 
transitions:
f0(0) → 0
+0(0, 0) → 1
+1(0, 0) → 2
f1(2) → 1
f1(2) → 2