details

property	value
status	complete
benchmark	Ex1_GM99_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n035.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	28.6405940056 seconds
cpu usage	87.123949747
max memory	1.1640328192E10

stage attributes

key	value
output-size	5481
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 7] active(f(active(c),mark(b),active(c))) -> active(f(active(c),mark(b),active(c))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {}. We have r|p = active(f(active(c),mark(b),active(c))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = active(f(active(c),mark(b),active(c))) 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... ## 2 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [2 DP problems]: ## DP problem: Dependency pairs = [active^#(f(a,b,_0)) -> mark^#(f(_0,_0,_0)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,_1,mark(_2))), mark^#(f(_0,_1,_2)) -> mark^#(_2)] TRS = {active(f(a,b,_0)) -> mark(f(_0,_0,_0)), active(c) -> mark(a), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,_1,mark(_2))), mark(a) -> active(a), mark(b) -> active(b), mark(c) -> active(c), f(mark(_0),_1,_2) -> f(_0,_1,_2), f(_0,mark(_1),_2) -> f(_0,_1,_2), f(_0,_1,mark(_2)) -> f(_0,_1,_2), f(active(_0),_1,_2) -> f(_0,_1,_2), f(_0,active(_1),_2) -> f(_0,_1,_2), f(_0,_1,active(_2)) -> f(_0,_1,_2)} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! Don't know whether this DP problem is finite. ## DP problem: Dependency pairs = [f^#(mark(_0),_1,_2) -> f^#(_0,_1,_2), f^#(_0,mark(_1),_2) -> f^#(_0,_1,_2), f^#(_0,_1,mark(_2)) -> f^#(_0,_1,_2), f^#(active(_0),_1,_2) -> f^#(_0,_1,_2), f^#(_0,active(_1),_2) -> f^#(_0,_1,_2), f^#(_0,_1,active(_2)) -> f^#(_0,_1,_2)] TRS = {active(f(a,b,_0)) -> mark(f(_0,_0,_0)), active(c) -> mark(a), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,_1,mark(_2))), mark(a) -> active(a), mark(b) -> active(b), mark(c) -> active(c), f(mark(_0),_1,_2) -> f(_0,_1,_2), f(_0,mark(_1),_2) -> f(_0,_1,_2), f(_0,_1,mark(_2)) -> f(_0,_1,_2), f(active(_0),_1,_2) -> f(_0,_1,_2), f(_0,active(_1),_2) -> f(_0,_1,_2), f(_0,_1,active(_2)) -> f(_0,_1,_2)} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## A DP problem could not be proved finite. ## Now, we try to prove that this problem is infinite. ## Trying to find a loop (forward=true, backward=true, max=20) # max_depth=20, unfold_variables=false: # Iteration 0: no loop found, 3 unfolded rules generated. # Iteration 1: no loop found, 8 unfolded rules generated. # Iteration 2: no loop found, 54 unfolded rules generated. # Iteration 3: no loop found, 534 unfolded rules generated. # Iteration 4: no loop found, 5972 unfolded rules generated. # Iteration 5: no loop found, 71470 unfolded rules generated. # Iteration 6: no loop found, 888619 unfolded rules generated. # Iteration 7: success, found a loop, 47165 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = active^#(f(a,b,_0)) -> mark^#(f(_0,_0,_0)) [trans] is in U_IR^0. D = mark^#(f(_0,_1,_2)) -> active^#(f(_0,_1,mark(_2))) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = active^#(f(a,b,_0)) -> active^#(f(_0,_0,mark(_0))) [trans] is in U_IR^1. We build a unit triple from L1. ==> L2 = active^#(f(a,b,_0)) -> active^#(f(_0,_0,mark(_0))) [unit] is in U_IR^2. Let p2 = [0]. We unfold the rule of L2 forwards at position p2 with the rule f(_0,_1,mark(_2)) -> f(_0,_1,_2). ==> L3 = active^#(f(a,b,_0)) -> active^#(f(_0,_0,_0)) [unit] is in U_IR^3. Let p3 = [0]. We unfold the rule of L3 backwards at position p3 with the rule f(mark(_0),_1,_2) -> f(_0,_1,_2). ==> L4 = active^#(f(mark(a),b,_0)) -> active^#(f(_0,_0,_0)) [unit] is in U_IR^4. Let p4 = [0, 0]. We unfold the rule of L4 backwards at position p4 with the rule active(c) -> mark(a). ==> L5 = active^#(f(active(c),b,active(c))) -> active^#(f(active(c),active(c),active(c))) [unit] is in U_IR^5. Let p5 = [0]. We unfold the rule of L5 backwards at position p5 with the rule f(_0,mark(_1),_2) -> f(_0,_1,_2). ==> L6 = active^#(f(active(c),mark(b),active(c))) -> active^#(f(active(c),active(c),active(c))) [unit] is in U_IR^6. Let p6 = [0, 1]. We unfold the rule of L6 forwards at position p6 with the rule active(c) -> mark(b). ==> L7 = active^#(f(active(c),mark(b),active(c))) -> active^#(f(active(c),mark(b),active(c))) [unit] is in U_IR^7. This DP problem is infinite. Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64 using Java version 1.8.0_292 ** END proof description ** Total number of generated unfolded rules = 1422679 popout

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actions

all output return to TRS Standard