NO

Prover = TRS(tech=GUIDED_UNF, nb_unfoldings=unlimited, unfold_variables=true, strategy=LEFTMOST_NE)

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 5] f(u(d,h(d),_0),k(b),u(d,h(d),_0)) -> f(u(d,h(d),_0),k(b),u(d,h(d),_0))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {}.
We have r|p = f(u(d,h(d),_0),k(b),u(d,h(d),_0)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(u(d,h(d),_0),k(b),u(d,h(d),_0)) loops w.r.t. the analyzed TRS.
** END proof argument **

** BEGIN proof description **
## Searching for a generalized rewrite rule
(a rule whose right-hand side contains a variable
that does not occur in the left-hand side)...
No generalized rewrite rule found!

## Searching for a loop by unfolding (unfolding of variable subterms: ON)...
# Iteration 0: no loop detected, 1 unfolded rule generated.
# Iteration 1: no loop detected, 2 unfolded rules generated.
# Iteration 2: no loop detected, 3 unfolded rules generated.
# Iteration 3: no loop detected, 3 unfolded rules generated.
# Iteration 4: no loop detected, 9 unfolded rules generated.
# Iteration 5: loop detected, 7 unfolded rules generated.
Here is the successful unfolding. Let IR be the TRS under analysis.
L0 = f^#(k(a),k(b),_0) -> f^#(_0,_0,_0) is in U_IR^0.
Let p0 = [0].
We unfold the rule of L0 backwards at position p0
with the rule u(d,c(_0),_1) -> k(_0).
==> L1 = f^#(u(d,c(a),_0),k(b),_1) -> f^#(_1,_1,_1) is in U_IR^1.
Let p1 = [0, 1].
We unfold the rule of L1 backwards at position p1
with the rule h(d) -> c(a).
==> L2 = f^#(u(d,h(d),_0),k(b),_1) -> f^#(_1,_1,_1) is in U_IR^2.
Let p2 = [0].
The subterm at position p2 in the left-hand side of the rule of L2 unifies with
the subterm at position p2 in the right-hand side of the rule of L2.
==> L3 = f^#(u(d,h(d),_0),k(b),u(d,h(d),_0)) -> f^#(u(d,h(d),_0),u(d,h(d),_0),u(d,h(d),_0)) is in U_IR^3.
Let p3 = [1, 1].
We unfold the rule of L3 forwards at position p3
with the rule h(d) -> c(b).
==> L4 = f^#(u(d,h(d),_0),k(b),u(d,h(d),_0)) -> f^#(u(d,h(d),_0),u(d,c(b),_0),u(d,h(d),_0)) is in U_IR^4.
Let p4 = [1].
We unfold the rule of L4 forwards at position p4
with the rule u(d,c(_0),_1) -> k(_0).
==> L5 = f^#(u(d,h(d),_0),k(b),u(d,h(d),_0)) -> f^#(u(d,h(d),_0),k(b),u(d,h(d),_0)) is in U_IR^5.

** END proof description **

Proof stopped at iteration 5
Number of unfolded rules generated by this proof = 25
Number of unfolded rules generated by all the parallel proofs = 25