Left Termination of the query pattern reverse_in_3(g, a, g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
↳3 PrologToPiTRSProof (⇐)
↳4 PiTRS
↳5 DependencyPairsProof (⇔)
↳6 PiDP
↳7 DependencyGraphProof (⇔)
↳8 PiDP
↳9 UsableRulesProof (⇔)
↳10 PiDP
↳11 PiDPToQDPProof (⇐)
↳12 QDP
↳13 QDPSizeChangeProof (⇔)
↳14 YES

(0) Obligation:

Clauses:

reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).

Queries:

reverse(g,a,g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

reverse1([], T5, T5).
reverse1(.(T28, []), .(T28, T29), T29).
reverse1(.(T43, .(T40, T41)), T45, T44) :- reverse1(T41, T45, .(T40, .(T43, T44))).

Queries:

reverse1(g,a,g).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse1_in: (b,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse1_in_gag([], T5, T5) → reverse1_out_gag([], T5, T5)
reverse1_in_gag(.(T28, []), .(T28, T29), T29) → reverse1_out_gag(.(T28, []), .(T28, T29), T29)
reverse1_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverse1_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverse1_out_gag(.(T43, .(T40, T41)), T45, T44)

The argument filtering Pi contains the following mapping:
reverse1_in_gag(x1, x2, x3)  =  reverse1_in_gag(x1, x3)
[]  =  []
reverse1_out_gag(x1, x2, x3)  =  reverse1_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse1_in_gag([], T5, T5) → reverse1_out_gag([], T5, T5)
reverse1_in_gag(.(T28, []), .(T28, T29), T29) → reverse1_out_gag(.(T28, []), .(T28, T29), T29)
reverse1_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverse1_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverse1_out_gag(.(T43, .(T40, T41)), T45, T44)

The argument filtering Pi contains the following mapping:
reverse1_in_gag(x1, x2, x3)  =  reverse1_in_gag(x1, x3)
[]  =  []
reverse1_out_gag(x1, x2, x3)  =  reverse1_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REVERSE1_IN_GAG(.(T43, .(T40, T41)), T45, T44) → U1_GAG(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
REVERSE1_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSE1_IN_GAG(T41, T45, .(T40, .(T43, T44)))

The TRS R consists of the following rules:

reverse1_in_gag([], T5, T5) → reverse1_out_gag([], T5, T5)
reverse1_in_gag(.(T28, []), .(T28, T29), T29) → reverse1_out_gag(.(T28, []), .(T28, T29), T29)
reverse1_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverse1_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverse1_out_gag(.(T43, .(T40, T41)), T45, T44)

The argument filtering Pi contains the following mapping:
reverse1_in_gag(x1, x2, x3)  =  reverse1_in_gag(x1, x3)
[]  =  []
reverse1_out_gag(x1, x2, x3)  =  reverse1_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)
REVERSE1_IN_GAG(x1, x2, x3)  =  REVERSE1_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6)  =  U1_GAG(x1, x2, x3, x5, x6)

We have to consider all (P,R,Pi)-chains

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE1_IN_GAG(.(T43, .(T40, T41)), T45, T44) → U1_GAG(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
REVERSE1_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSE1_IN_GAG(T41, T45, .(T40, .(T43, T44)))

The TRS R consists of the following rules:

reverse1_in_gag([], T5, T5) → reverse1_out_gag([], T5, T5)
reverse1_in_gag(.(T28, []), .(T28, T29), T29) → reverse1_out_gag(.(T28, []), .(T28, T29), T29)
reverse1_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverse1_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverse1_out_gag(.(T43, .(T40, T41)), T45, T44)

The argument filtering Pi contains the following mapping:
reverse1_in_gag(x1, x2, x3)  =  reverse1_in_gag(x1, x3)
[]  =  []
reverse1_out_gag(x1, x2, x3)  =  reverse1_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)
REVERSE1_IN_GAG(x1, x2, x3)  =  REVERSE1_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6)  =  U1_GAG(x1, x2, x3, x5, x6)

We have to consider all (P,R,Pi)-chains

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(8) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE1_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSE1_IN_GAG(T41, T45, .(T40, .(T43, T44)))

The TRS R consists of the following rules:

reverse1_in_gag([], T5, T5) → reverse1_out_gag([], T5, T5)
reverse1_in_gag(.(T28, []), .(T28, T29), T29) → reverse1_out_gag(.(T28, []), .(T28, T29), T29)
reverse1_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverse1_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverse1_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverse1_out_gag(.(T43, .(T40, T41)), T45, T44)

The argument filtering Pi contains the following mapping:
reverse1_in_gag(x1, x2, x3)  =  reverse1_in_gag(x1, x3)
[]  =  []
reverse1_out_gag(x1, x2, x3)  =  reverse1_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)
REVERSE1_IN_GAG(x1, x2, x3)  =  REVERSE1_IN_GAG(x1, x3)

We have to consider all (P,R,Pi)-chains

(9) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(10) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

REVERSE1_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSE1_IN_GAG(T41, T45, .(T40, .(T43, T44)))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
REVERSE1_IN_GAG(x1, x2, x3)  =  REVERSE1_IN_GAG(x1, x3)

We have to consider all (P,R,Pi)-chains

(11) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REVERSE1_IN_GAG(.(T43, .(T40, T41)), T44) → REVERSE1_IN_GAG(T41, .(T40, .(T43, T44)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • REVERSE1_IN_GAG(.(T43, .(T40, T41)), T44) → REVERSE1_IN_GAG(T41, .(T40, .(T43, T44)))
    The graph contains the following edges 1 > 1

(14) YES