NO Termination proof of assoc.trs

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

f(a, x) → f(x, f(b, x))

The relative TRS consists of the following S rules:

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

(1) RelTRSLoopFinderProof (COMPLETE transformation)

The following loop was found:

---------- Loop: ----------

f(a, f(x', f(y', a))) → f(a, f(f(x', y'), a)) with rule f(x'', f(y'', z)) → f(f(x'', y''), z) at position [1] and matcher [x'' / x', y'' / y', z / a]

f(a, f(f(x', y'), a)) → f(f(f(x', y'), a), f(b, f(f(x', y'), a))) with rule f(a, x) → f(x, f(b, x)) at position [] and matcher [x / f(f(x', y'), a)]

f(f(f(x', y'), a), f(b, f(f(x', y'), a))) → f(f(x', y'), f(a, f(b, f(f(x', y'), a)))) with rule f(f(x, y), z) → f(x, f(y, z)) at position [] and matcher [x / f(x', y'), y / a, z / f(b, f(f(x', y'), a))]

Now an instance of the first term with Matcher [x' / b, y' / f(x', y')] occurs in the last term at position [1].

Context: f(f(x', y'), [])

Therefore, the relative TRS problem does not terminate.

(2) NO