(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)).
Query: som3(g,a,a)
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. 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) YES