### (0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, 1).

The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, z) → 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:
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))

### (2) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, 1).

The TRS R consists of the following rules:

b(y, z) → z
f(c(a, z, x)) → b(a, z)

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, 1).

The TRS R consists of the following rules:

b(y, z) → z
f(c(a, z, x)) → b(a, z)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

b(z0, z1) → z1
f(c(a, z0, z1)) → b(a, z0)
Tuples:

B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
S tuples:

B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
K tuples:none
Defined Rule Symbols:

b, f

Defined Pair Symbols:

B, F

Compound Symbols:

c1, c2

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

Removed 2 trailing nodes:

B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

b(z0, z1) → z1
f(c(a, z0, z1)) → b(a, z0)
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

b, f

Defined Pair Symbols:none

Compound Symbols:none

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

The set S is empty