Left Termination of the query pattern som3_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
    ↳11 QDPSizeChangeProof (⇔)
    ↳12 TRUE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇐)
    ↳22 QDP

(0) Obligation:

Clauses:

som3([], Bs, Bs).
som3(As, [], As).
som3(.(A, As), .(B, Bs), .(+(A, B), Cs)) :- som3(As, Bs, Cs).
som4_1(As, Bs, Cs, Ds) :- ','(som3(As, Bs, Es), som3(Es, Cs, Ds)).
som4_2(As, Bs, Cs, Ds) :- ','(som3(Es, Cs, Ds), som3(As, Bs, Es)).

Queries:

som3(g,a,a).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
som3_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)

(3) 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:

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)

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

(4) Obligation:

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

SOM3_IN_GAA(.(A, As)) → SOM3_IN_GAA(As)

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

(11) 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:

  • SOM3_IN_GAA(.(A, As)) → SOM3_IN_GAA(As)
    The graph contains the following edges 1 > 1

(12) TRUE

(13) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
som3_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

(15) 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:

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x6)

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

(16) Obligation:

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x6)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) Obligation:

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)

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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) Obligation:

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

SOM3_IN_GAA(.(A, As)) → SOM3_IN_GAA(As)

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