details

property	value
status	complete
benchmark	Ex6_15_AEL02_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	0.304502 seconds
cpu usage	0.638322
user time	0.579365
system time	0.058957
max virtual memory	113188.0
max residence set size	86748.0

stage attributes

key	value
starexec-result	NO

output

NO Prover = TRS(tech=GUIDED_UNF_TRIPLES, nb_unfoldings=unlimited, unfold_variables=false, max_nb_coefficients=12, max_nb_unfolded_rules=-1, strategy=LEFTMOST_NE) ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 4] quote1(n__cons(_0,n__from(_1))) -> quote1(n__cons(_1,n__from(s(_1)))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {_2->_1} and theta2 = {_2->s(_2), _0->_2}. We have r|p = quote1(n__cons(_1,n__from(s(_1)))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = quote1(n__cons(_0,n__from(_1))) loops w.r.t. the analyzed TRS. ** END proof argument ** ** BEGIN proof description ** ## Searching for a generalized rewrite rule (a rule whose right-hand side contains a variable that does not occur in the left-hand side)... No generalized rewrite rule found! ## Applying the DP framework... ## Round 1: ## DP problem: Dependency pairs = [unquote1^#(cons1(_0,_1)) -> unquote1^#(_1)] TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, s(_0) -> n__s(_0), sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(_0,_1), activate(n__from(_0)) -> from(_0), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(_0,_1), activate(n__nil) -> nil, activate(n__s(_0)) -> s(_0), activate(n__sel(_0,_1)) -> sel(_0,_1), activate(_0) -> _0} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [unquote^#(s1(_0)) -> unquote^#(_0)] TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, s(_0) -> n__s(_0), sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(_0,_1), activate(n__from(_0)) -> from(_0), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(_0,_1), activate(n__nil) -> nil, activate(n__s(_0)) -> s(_0), activate(n__sel(_0,_1)) -> sel(_0,_1), activate(_0) -> _0} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [quote1^#(n__cons(_0,_1)) -> quote1^#(activate(_1))] TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, s(_0) -> n__s(_0), sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(_0,_1), activate(n__from(_0)) -> from(_0), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(_0,_1), activate(n__nil) -> nil, activate(n__s(_0)) -> s(_0), activate(n__sel(_0,_1)) -> sel(_0,_1), activate(_0) -> _0} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (67)! Aborting! ## Trying with lexicographic path orders... Too many argument filtering possibilities (60466176)! Aborting! ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS # max_depth=3, unfold_variables=false: # Iteration 0: nontermination not detected, 1 unfolded rule generated. # Iteration 1: nontermination not detected, 1 unfolded rule generated. # Iteration 2: nontermination not detected, 4 unfolded rules generated. # Iteration 3: nontermination not detected, 0 unfolded rule generated. Nontermination not detected! # max_depth=3, unfold_variables=true: # Iteration 0: nontermination not detected, 1 unfolded rule generated. # Iteration 1: nontermination not detected, 1 unfolded rule generated. # Iteration 2: nontermination not detected, 4 unfolded rules generated. # Iteration 3: nontermination not detected, 0 unfolded rule generated. Nontermination not detected! # max_depth=4, unfold_variables=false: # Iteration 0: nontermination not detected, 1 unfolded rule generated. # Iteration 1: nontermination not detected, 1 unfolded rule generated. # Iteration 2: nontermination not detected, 4 unfolded rules generated. # Iteration 3: nontermination not detected, 3 unfolded rules generated. # Iteration 4: nontermination detected, 5 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = quote1^#(n__cons(_0,_1)) -> quote1^#(activate(_1)) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = quote1^#(n__cons(_0,_1)) -> quote1^#(activate(_1)) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 forwards at position p1 with the rule activate(n__from(_0)) -> from(_0). ==> L2 = quote1^#(n__cons(_0,n__from(_1))) -> quote1^#(from(_1)) [unit] is in U_IR^2. Let p2 = [0]. We unfold the rule of L2 forwards at position p2 with the rule from(_0) -> cons(_0,n__from(s(_0))). ==> L3 = quote1^#(n__cons(_0,n__from(_1))) -> quote1^#(cons(_1,n__from(s(_1)))) [unit] is in U_IR^3. Let p3 = [0]. We unfold the rule of L3 forwards at position p3 with the rule cons(_0,_1) -> n__cons(_0,_1). ==> L4 = quote1^#(n__cons(_0,n__from(_1))) -> quote1^#(n__cons(_1,n__from(s(_1)))) [unit] is in U_IR^4. This DP problem is infinite. ** END proof description ** Proof stopped at iteration 4 Number of unfolded rules generated by this proof = 26 Number of unfolded rules generated by all the parallel proofs = 194 popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Standard