details

property	value
status	complete
benchmark	thiemann26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	1.64341 seconds
cpu usage	4.92988
user time	4.70708
system time	0.2228
max virtual memory	3.5399428E7
max residence set size	622616.0

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 = 7] if(ge(id_inc(0),s(0)),true,_0,id_inc(0),_1) -> if(ge(id_inc(0),s(0)),true,minus(_0,id_inc(0)),id_inc(0),id_inc(_1)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_1->id_inc(_1), _0->minus(_0,id_inc(0))}. We have r|p = if(ge(id_inc(0),s(0)),true,minus(_0,id_inc(0)),id_inc(0),id_inc(_1)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = if(ge(id_inc(0),s(0)),true,_0,id_inc(0),_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 = [div^#(_0,_1,_2) -> if^#(ge(_1,s(0)),ge(_0,_1),_0,_1,_2), if^#(true,true,_0,_1,_2) -> div^#(minus(_0,_1),_1,id_inc(_2))] TRS = {ge(_0,0) -> true, ge(0,s(_0)) -> false, ge(s(_0),s(_1)) -> ge(_0,_1), minus(_0,0) -> _0, minus(0,_0) -> 0, minus(s(_0),s(_1)) -> minus(_0,_1), id_inc(_0) -> _0, id_inc(_0) -> s(_0), quot(_0,_1) -> div(_0,_1,0), div(_0,_1,_2) -> if(ge(_1,s(0)),ge(_0,_1),_0,_1,_2), if(false,_0,_1,_2,_3) -> div_by_zero, if(true,false,_0,_1,_2) -> _2, if(true,true,_0,_1,_2) -> div(minus(_0,_1),_1,id_inc(_2))} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (61)! Aborting! ## Trying with lexicographic path orders... Failed! ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS # max_depth=20, unfold_variables=false: # Iteration 0: nontermination not detected, 2 unfolded rules generated. # Iteration 1: nontermination not detected, 3 unfolded rules generated. # Iteration 2: nontermination not detected, 10 unfolded rules generated. # Iteration 3: nontermination not detected, 2 unfolded rules generated. # Iteration 4: nontermination not detected, 8 unfolded rules generated. # Iteration 5: nontermination not detected, 34 unfolded rules generated. # Iteration 6: nontermination not detected, 201 unfolded rules generated. # Iteration 7: nontermination detected, 1150 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = if^#(true,true,_0,_1,_2) -> div^#(minus(_0,_1),_1,id_inc(_2)) [trans] is in U_IR^0. D = div^#(_0,_1,_2) -> if^#(ge(_1,s(0)),ge(_0,_1),_0,_1,_2) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = if^#(true,true,_0,_1,_2) -> if^#(ge(_1,s(0)),ge(minus(_0,_1),_1),minus(_0,_1),_1,id_inc(_2)) [trans] is in U_IR^1. We build a unit triple from L1. ==> L2 = if^#(true,true,_0,_1,_2) -> if^#(ge(_1,s(0)),ge(minus(_0,_1),_1),minus(_0,_1),_1,id_inc(_2)) [unit] is in U_IR^2. Let p2 = [0]. We unfold the rule of L2 backwards at position p2 with the rule ge(_0,0) -> true. ==> L3 = if^#(ge(_0,0),true,_1,_2,_3) -> if^#(ge(_2,s(0)),ge(minus(_1,_2),_2),minus(_1,_2),_2,id_inc(_3)) [unit] is in U_IR^3. Let p3 = [0]. We unfold the rule of L3 backwards at position p3 with the rule ge(s(_0),s(_1)) -> ge(_0,_1). ==> L4 = if^#(ge(s(_0),s(0)),true,_1,_2,_3) -> if^#(ge(_2,s(0)),ge(minus(_1,_2),_2),minus(_1,_2),_2,id_inc(_3)) [unit] is in U_IR^4. Let p4 = [0, 0]. We unfold the rule of L4 backwards at position p4 with the rule id_inc(_0) -> s(_0). ==> L5 = if^#(ge(id_inc(_0),s(0)),true,_1,id_inc(_0),_2) -> if^#(ge(id_inc(_0),s(0)),ge(minus(_1,id_inc(_0)),id_inc(_0)),minus(_1,id_inc(_0)),id_inc(_0),id_inc(_2)) [unit] is in U_IR^5. Let p5 = [1, 1]. We unfold the rule of L5 forwards at position p5 with the rule id_inc(_0) -> _0. ==> L6 = if^#(ge(id_inc(_0),s(0)),true,_1,id_inc(_0),_2) -> if^#(ge(id_inc(_0),s(0)),ge(minus(_1,id_inc(_0)),_0),minus(_1,id_inc(_0)),id_inc(_0),id_inc(_2)) [unit] is in U_IR^6. Let p6 = [1]. We unfold the rule of L6 forwards at position p6 with the rule ge(_0,0) -> true. ==> L7 = if^#(ge(id_inc(0),s(0)),true,_0,id_inc(0),_1) -> if^#(ge(id_inc(0),s(0)),true,minus(_0,id_inc(0)),id_inc(0),id_inc(_1)) [unit] is in U_IR^7. This DP problem is infinite. ** END proof description ** Proof stopped at iteration 7 Number of unfolded rules generated by this proof = 1410 Number of unfolded rules generated by all the parallel proofs = 5346 popout

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