NO Termination proof of rt3-8.trs

(0) Obligation:

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

f(s(x), y) → f(-(s(x), y), y)

The relative TRS consists of the following S rules:

+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
f(x, y) → f(x, +(x, y))
+(0, y) → y

(1) RelTRSLoopFinderProof (COMPLETE transformation)

The following loop was found:

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

f(-(s(x), 0), 0) → f(s(x), 0) with rule -(x', 0) → x' at position [0] and matcher [x' / s(x)]

f(s(x), 0) → f(-(s(x), 0), 0) with rule f(s(x'), y) → f(-(s(x'), y), y) at position [] and matcher [x' / x, y / 0]

Now an instance of the first term with Matcher [ ] occurs in the last term at position [].

Context: []

Therefore, the relative TRS problem does not terminate.

(2) NO