Termination of the following Term Rewriting System could be disproven:
Generalized rewrite system (where rules with free variables on rhs are allowed):
The TRS R consists of the following rules:
sel(0, cons) → X
sel(s, cons) → sel(X, Z)
dbl1(s) → s1(s1(dbl1(X)))
sel1(0, cons) → X
sel1(s, cons) → sel1(X, Z)
quote(s) → s1(quote(X))
dbl(0) → 0
dbl(s) → s
dbls(nil) → nil
dbls(cons) → cons
indx(nil) → nil
indx(cons) → cons
from → cons
dbl1(0) → 01
quote(0) → 01
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
↳ GTRS
↳ CritRuleProof
Generalized rewrite system (where rules with free variables on rhs are allowed):
The TRS R consists of the following rules:
sel(0, cons) → X
sel(s, cons) → sel(X, Z)
dbl1(s) → s1(s1(dbl1(X)))
sel1(0, cons) → X
sel1(s, cons) → sel1(X, Z)
quote(s) → s1(quote(X))
dbl(0) → 0
dbl(s) → s
dbls(nil) → nil
dbls(cons) → cons
indx(nil) → nil
indx(cons) → cons
from → cons
dbl1(0) → 01
quote(0) → 01
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
The rule sel(0, cons) → X contains free variables in its right-hand side. Hence the TRS is not-terminating.