Left Termination of the query pattern
som3_in_3(g, a, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
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).
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:
som3_in(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
som3_in(As, [], As) → som3_out(As, [], As)
som3_in([], Bs, Bs) → som3_out([], Bs, Bs)
U1(A, As, B, Bs, Cs, som3_out(As, Bs, Cs)) → som3_out(.(A, As), .(B, Bs), .(+(A, B), Cs))
The argument filtering Pi contains the following mapping:
som3_in(x1, x2, x3) = som3_in(x1)
.(x1, x2) = .(x1, x2)
+(x1, x2) = +(x1, x2)
U1(x1, x2, x3, x4, x5, x6) = U1(x6)
[] = []
som3_out(x1, x2, x3) = som3_out
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
som3_in(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
som3_in(As, [], As) → som3_out(As, [], As)
som3_in([], Bs, Bs) → som3_out([], Bs, Bs)
U1(A, As, B, Bs, Cs, som3_out(As, Bs, Cs)) → som3_out(.(A, As), .(B, Bs), .(+(A, B), Cs))
The argument filtering Pi contains the following mapping:
som3_in(x1, x2, x3) = som3_in(x1)
.(x1, x2) = .(x1, x2)
+(x1, x2) = +(x1, x2)
U1(x1, x2, x3, x4, x5, x6) = U1(x6)
[] = []
som3_out(x1, x2, x3) = som3_out
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
SOM3_IN(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U11(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
SOM3_IN(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN(As, Bs, Cs)
The TRS R consists of the following rules:
som3_in(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
som3_in(As, [], As) → som3_out(As, [], As)
som3_in([], Bs, Bs) → som3_out([], Bs, Bs)
U1(A, As, B, Bs, Cs, som3_out(As, Bs, Cs)) → som3_out(.(A, As), .(B, Bs), .(+(A, B), Cs))
The argument filtering Pi contains the following mapping:
som3_in(x1, x2, x3) = som3_in(x1)
.(x1, x2) = .(x1, x2)
+(x1, x2) = +(x1, x2)
U1(x1, x2, x3, x4, x5, x6) = U1(x6)
[] = []
som3_out(x1, x2, x3) = som3_out
SOM3_IN(x1, x2, x3) = SOM3_IN(x1)
U11(x1, x2, x3, x4, x5, x6) = U11(x6)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
SOM3_IN(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U11(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
SOM3_IN(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN(As, Bs, Cs)
The TRS R consists of the following rules:
som3_in(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
som3_in(As, [], As) → som3_out(As, [], As)
som3_in([], Bs, Bs) → som3_out([], Bs, Bs)
U1(A, As, B, Bs, Cs, som3_out(As, Bs, Cs)) → som3_out(.(A, As), .(B, Bs), .(+(A, B), Cs))
The argument filtering Pi contains the following mapping:
som3_in(x1, x2, x3) = som3_in(x1)
.(x1, x2) = .(x1, x2)
+(x1, x2) = +(x1, x2)
U1(x1, x2, x3, x4, x5, x6) = U1(x6)
[] = []
som3_out(x1, x2, x3) = som3_out
SOM3_IN(x1, x2, x3) = SOM3_IN(x1)
U11(x1, x2, x3, x4, x5, x6) = U11(x6)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
SOM3_IN(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN(As, Bs, Cs)
The TRS R consists of the following rules:
som3_in(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1(A, As, B, Bs, Cs, som3_in(As, Bs, Cs))
som3_in(As, [], As) → som3_out(As, [], As)
som3_in([], Bs, Bs) → som3_out([], Bs, Bs)
U1(A, As, B, Bs, Cs, som3_out(As, Bs, Cs)) → som3_out(.(A, As), .(B, Bs), .(+(A, B), Cs))
The argument filtering Pi contains the following mapping:
som3_in(x1, x2, x3) = som3_in(x1)
.(x1, x2) = .(x1, x2)
+(x1, x2) = +(x1, x2)
U1(x1, x2, x3, x4, x5, x6) = U1(x6)
[] = []
som3_out(x1, x2, x3) = som3_out
SOM3_IN(x1, x2, x3) = SOM3_IN(x1)
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
Pi DP problem:
The TRS P consists of the following rules:
SOM3_IN(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN(As, Bs, Cs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
+(x1, x2) = +(x1, x2)
SOM3_IN(x1, x2, x3) = SOM3_IN(x1)
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
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
SOM3_IN(.(A, As)) → SOM3_IN(As)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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(.(A, As)) → SOM3_IN(As)
The graph contains the following edges 1 > 1