/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 147 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   if(true, t, e) -> t
   if(false, t, e) -> e
   member(x, nil) -> false
   member(x, cons(y, ys)) -> if(eq(x, y), true, member(x, ys))
   eq(nil, nil) -> true
   eq(O(x), 0(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(1(x), 1(y)) -> eq(x, y)
   negate(0(x)) -> 1(x)
   negate(1(x)) -> 0(x)
   choice(cons(x, xs)) -> x
   choice(cons(x, xs)) -> choice(xs)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true
   verify(cons(l, ls)) -> if(member(negate(l), ls), false, verify(ls))
   sat(cnf) -> satck(cnf, guess(cnf))
   satck(cnf, assign) -> if(verify(assign), assign, unsat)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
if/3(YES,YES,YES)
true/0)
false/0)
member/2(YES,YES)
nil/0)
cons/2(YES,YES)
eq/2(YES,YES)
O/1)YES(
0/1(YES)
1/1)YES(
negate/1(YES)
choice/1)YES(
guess/1)YES(
verify/1)YES(
sat/1(YES)
satck/2(YES,YES)
unsat/0)

Quasi precedence:
[true, false, nil, cons_2, 0_1, negate_1] > member_2 > eq_2 > if_3
sat_1 > [satck_2, unsat] > if_3


Status:
if_3: multiset status
true: multiset status
false: multiset status
member_2: [1,2]
nil: multiset status
cons_2: [1,2]
eq_2: [2,1]
0_1: [1]
negate_1: [1]
sat_1: [1]
satck_2: [2,1]
unsat: multiset status

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   if(true, t, e) -> t
   if(false, t, e) -> e
   member(x, nil) -> false
   member(x, cons(y, ys)) -> if(eq(x, y), true, member(x, ys))
   eq(nil, nil) -> true
   eq(O(x), 0(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   negate(0(x)) -> 1(x)
   choice(cons(x, xs)) -> x
   choice(cons(x, xs)) -> choice(xs)
   verify(cons(l, ls)) -> if(member(negate(l), ls), false, verify(ls))
   sat(cnf) -> satck(cnf, guess(cnf))
   satck(cnf, assign) -> if(verify(assign), assign, unsat)




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   eq(1(x), 1(y)) -> eq(x, y)
   negate(1(x)) -> 0(x)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:choice_1 > verify_1 > true > guess_1 > cons_2 > nil > negate_1 > 0_1 > 1_1 > eq_2

and weight map:

   nil=1
   true=2
   1_1=1
   negate_1=1
   0_1=2
   guess_1=1
   choice_1=0
   verify_1=1
   cons_2=0
   eq_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   eq(1(x), 1(y)) -> eq(x, y)
   negate(1(x)) -> 0(x)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(5) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES