YES
(VAR x y)
(RULES 
minus(x,0) -> x
minus(s(x),s(y)) -> minus(x,y)
quot(0,s(y)) -> 0
quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
)

The TRS is an overlay system and all critical pairs are trivial, thus termination of innermost rewriting is equivalent to termination of rewriting.

Proving termination of innermost rewriting for _3_1:

-> Dependency pairs:
nF_minus(s(x),s(y)) -> nF_minus(x,y)
nF_quot(s(x),s(y)) -> nF_quot(minus(x,y),s(y))
nF_quot(s(x),s(y)) -> nF_minus(x,y)

-> Proof of termination for _3_1_1_1:
-> -> Dependency pairs in cycle:
nF_quot(s(x),s(y)) -> nF_quot(minus(x,y),s(y))

UsableRules:
minus(x,0) -> x
minus(s(x),s(y)) -> minus(x,y)

Polynomial Interpretation:

[minus](X1,X2) = X1
[0] = 0
[s](X) = X + 1
[quot](X1,X2) = 0
[nF_quot](X1,X2) = X1

TIME: 4.3598e-2


-> Proof of termination for _3_1_1_2:
-> -> Dependency pairs in cycle:
nF_minus(s(x),s(y)) -> nF_minus(x,y)

Termination proved: Cycles verify subterm criterion.

SETTINGS:
Base ordering: Polynomial ordering
Proof mode: SCCs in DG + base ordering
Upper bound for coeffs: 1
Rationals below 1 for all non-replacing args: No
Polynomial interpretation: Linear
Coeffs in polynomials: No rationals
Delta: automatic



Termination was proved succesfully.