Left Termination of the query pattern
reverse_in_2(a, g)
w.r.t. the given Prolog program could not be shown:
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
Clauses:
reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).
Queries:
reverse(a,g).
We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
reverse_out(x1, x2, x3) = reverse_out(x1)
reverse_out(x1, x2) = reverse_out(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PrologToPiTRSProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
reverse_out(x1, x2, x3) = reverse_out(x1)
reverse_out(x1, x2) = reverse_out(x1)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(X1s, X2s) → U11(X1s, X2s, reverse_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE_IN(X1s, [], X2s)
REVERSE_IN(.(X, X1s), X2s, Ys) → U21(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
reverse_out(x1, x2, x3) = reverse_out(x1)
reverse_out(x1, x2) = reverse_out(x1)
REVERSE_IN(x1, x2) = REVERSE_IN(x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
U21(x1, x2, x3, x4, x5) = U21(x5)
U11(x1, x2, x3) = U11(x3)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(X1s, X2s) → U11(X1s, X2s, reverse_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE_IN(X1s, [], X2s)
REVERSE_IN(.(X, X1s), X2s, Ys) → U21(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
reverse_out(x1, x2, x3) = reverse_out(x1)
reverse_out(x1, x2) = reverse_out(x1)
REVERSE_IN(x1, x2) = REVERSE_IN(x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
U21(x1, x2, x3, x4, x5) = U21(x5)
U11(x1, x2, x3) = U11(x3)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
reverse_out(x1, x2, x3) = reverse_out(x1)
reverse_out(x1, x2) = reverse_out(x1)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ Instantiation
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN(X2s, Ys) → REVERSE_IN(.(X2s), Ys)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN(X2s, Ys) → REVERSE_IN(.(X2s), Ys) we obtained the following new rules:
REVERSE_IN(.(z0), z1) → REVERSE_IN(.(.(z0)), z1)
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(z0), z1) → REVERSE_IN(.(.(z0)), z1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN(.(z0), z1) → REVERSE_IN(.(.(z0)), z1) we obtained the following new rules:
REVERSE_IN(.(.(z0)), z1) → REVERSE_IN(.(.(.(z0))), z1)
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ NonTerminationProof
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(.(z0)), z1) → REVERSE_IN(.(.(.(z0))), z1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
The TRS P consists of the following rules:
REVERSE_IN(.(.(z0)), z1) → REVERSE_IN(.(.(.(z0))), z1)
The TRS R consists of the following rules:none
s = REVERSE_IN(.(.(z0)), z1) evaluates to t =REVERSE_IN(.(.(.(z0))), z1)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [z0 / .(z0)]
- Semiunifier: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from REVERSE_IN(.(.(z0)), z1) to REVERSE_IN(.(.(.(z0))), z1).
We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x3, x4, x5)
reverse_out(x1, x2, x3) = reverse_out(x1, x2, x3)
reverse_out(x1, x2) = reverse_out(x1, x2)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x3, x4, x5)
reverse_out(x1, x2, x3) = reverse_out(x1, x2, x3)
reverse_out(x1, x2) = reverse_out(x1, x2)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(X1s, X2s) → U11(X1s, X2s, reverse_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE_IN(X1s, [], X2s)
REVERSE_IN(.(X, X1s), X2s, Ys) → U21(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x3, x4, x5)
reverse_out(x1, x2, x3) = reverse_out(x1, x2, x3)
reverse_out(x1, x2) = reverse_out(x1, x2)
REVERSE_IN(x1, x2) = REVERSE_IN(x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
U21(x1, x2, x3, x4, x5) = U21(x3, x4, x5)
U11(x1, x2, x3) = U11(x2, x3)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(X1s, X2s) → U11(X1s, X2s, reverse_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE_IN(X1s, [], X2s)
REVERSE_IN(.(X, X1s), X2s, Ys) → U21(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x3, x4, x5)
reverse_out(x1, x2, x3) = reverse_out(x1, x2, x3)
reverse_out(x1, x2) = reverse_out(x1, x2)
REVERSE_IN(x1, x2) = REVERSE_IN(x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
U21(x1, x2, x3, x4, x5) = U21(x3, x4, x5)
U11(x1, x2, x3) = U11(x2, x3)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
reverse_in(x1, x2, x3) = reverse_in(x2, x3)
[] = []
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4, x5) = U2(x3, x4, x5)
reverse_out(x1, x2, x3) = reverse_out(x1, x2, x3)
reverse_out(x1, x2) = reverse_out(x1, x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
REVERSE_IN(x1, x2, x3) = REVERSE_IN(x2, x3)
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN(X2s, Ys) → REVERSE_IN(.(X2s), Ys)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN(X2s, Ys) → REVERSE_IN(.(X2s), Ys) we obtained the following new rules:
REVERSE_IN(.(z0), z1) → REVERSE_IN(.(.(z0)), z1)
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(z0), z1) → REVERSE_IN(.(.(z0)), z1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN(.(z0), z1) → REVERSE_IN(.(.(z0)), z1) we obtained the following new rules:
REVERSE_IN(.(.(z0)), z1) → REVERSE_IN(.(.(.(z0))), z1)
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN(.(.(z0)), z1) → REVERSE_IN(.(.(.(z0))), z1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
The TRS P consists of the following rules:
REVERSE_IN(.(.(z0)), z1) → REVERSE_IN(.(.(.(z0))), z1)
The TRS R consists of the following rules:none
s = REVERSE_IN(.(.(z0)), z1) evaluates to t =REVERSE_IN(.(.(.(z0))), z1)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [z0 / .(z0)]
- Semiunifier: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from REVERSE_IN(.(.(z0)), z1) to REVERSE_IN(.(.(.(z0))), z1).