/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Left Termination of the query pattern

less(a,g)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) PiDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) PiDP
(9) PiDPToQDPProof [SOUND, 8 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) YES


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(0)
Obligation:
Clauses:

less(0, s(X1)).
less(s(X), s(Y)) :- less(X, Y).


Query: less(a,g)
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(1) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

less_in_2: (f,b)

Transforming Prolog into the following Term Rewriting System:

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

   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
   less_in_ag(s(X), s(Y)) -> U1_ag(X, Y, less_in_ag(X, Y))
   U1_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2)  =  less_in_ag(x2)

s(x1)  =  s(x1)

less_out_ag(x1, x2)  =  less_out_ag(x1)

U1_ag(x1, x2, x3)  =  U1_ag(x3)





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



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(2)
Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:

   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
   less_in_ag(s(X), s(Y)) -> U1_ag(X, Y, less_in_ag(X, Y))
   U1_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2)  =  less_in_ag(x2)

s(x1)  =  s(x1)

less_out_ag(x1, x2)  =  less_out_ag(x1)

U1_ag(x1, x2, x3)  =  U1_ag(x3)



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

   LESS_IN_AG(s(X), s(Y)) -> U1_AG(X, Y, less_in_ag(X, Y))
   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)

The TRS R consists of the following rules:

   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
   less_in_ag(s(X), s(Y)) -> U1_ag(X, Y, less_in_ag(X, Y))
   U1_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2)  =  less_in_ag(x2)

s(x1)  =  s(x1)

less_out_ag(x1, x2)  =  less_out_ag(x1)

U1_ag(x1, x2, x3)  =  U1_ag(x3)

LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)

U1_AG(x1, x2, x3)  =  U1_AG(x3)


We have to consider all (P,R,Pi)-chains
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(4)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   LESS_IN_AG(s(X), s(Y)) -> U1_AG(X, Y, less_in_ag(X, Y))
   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)

The TRS R consists of the following rules:

   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
   less_in_ag(s(X), s(Y)) -> U1_ag(X, Y, less_in_ag(X, Y))
   U1_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2)  =  less_in_ag(x2)

s(x1)  =  s(x1)

less_out_ag(x1, x2)  =  less_out_ag(x1)

U1_ag(x1, x2, x3)  =  U1_ag(x3)

LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)

U1_AG(x1, x2, x3)  =  U1_AG(x3)


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

(5) DependencyGraphProof (EQUIVALENT)
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:

   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)

The TRS R consists of the following rules:

   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
   less_in_ag(s(X), s(Y)) -> U1_ag(X, Y, less_in_ag(X, Y))
   U1_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2)  =  less_in_ag(x2)

s(x1)  =  s(x1)

less_out_ag(x1, x2)  =  less_out_ag(x1)

U1_ag(x1, x2, x3)  =  U1_ag(x3)

LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)


We have to consider all (P,R,Pi)-chains
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(7) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
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(8)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)

LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)


We have to consider all (P,R,Pi)-chains
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(9) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LESS_IN_AG(s(Y)) -> LESS_IN_AG(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
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(11) QDPSizeChangeProof (EQUIVALENT)
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:
*LESS_IN_AG(s(Y)) -> LESS_IN_AG(Y)
The graph contains the following edges 1 > 1


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(12)
YES