details

property	value
status	complete
benchmark	Ex24_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	13.5424909592 seconds
cpu usage	15.143712533
max memory	5.38918912E8

stage attributes

key	value
output-size	5685
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 = 6] active(f(mark(b),mark(c),c)) -> active(f(mark(b),mark(c),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(mark(b),mark(c),c)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = active(f(mark(b),mark(c),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(b,_0,c)) -> mark^#(f(_0,c,_0)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2)), mark^#(f(_0,_1,_2)) -> mark^#(_1)] TRS = {active(f(b,_0,c)) -> mark(f(_0,c,_0)), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,mark(_1),_2)), 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... Successfully decomposed the DP problem into 1 smaller problem to solve! ## 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(b,_0,c)) -> mark(f(_0,c,_0)), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,mark(_1),_2)), 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. ## Round 2 [1 DP problem]: ## DP problem: Dependency pairs = [active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2))] TRS = {active(f(b,_0,c)) -> mark(f(_0,c,_0)), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,mark(_1),_2)), 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... Failed! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! Don't know whether 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, 2 unfolded rules generated. # Iteration 1: no loop found, 4 unfolded rules generated. # Iteration 2: no loop found, 30 unfolded rules generated. # Iteration 3: no loop found, 12 unfolded rules generated. # Iteration 4: no loop found, 136 unfolded rules generated. # Iteration 5: no loop found, 1578 unfolded rules generated. # Iteration 6: success, found a loop, 3175 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)) [trans] is in U_IR^0. D = mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2)) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = active^#(f(b,_0,c)) -> active^#(f(_0,mark(c),_0)) [trans] is in U_IR^1. We build a unit triple from L1. ==> L2 = active^#(f(b,_0,c)) -> active^#(f(_0,mark(c),_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(b,mark(_0),c)) -> active^#(f(mark(_0),mark(c),_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(b),mark(_0),c)) -> active^#(f(mark(_0),mark(c),_0)) [unit] is in U_IR^4. Let p4 = [0, 0]. We unfold the rule of L4 forwards at position p4 with the rule mark(c) -> active(c). ==> L5 = active^#(f(mark(b),mark(c),c)) -> active^#(f(active(c),mark(c),c)) [unit] is in U_IR^5. Let p5 = [0, 0]. We unfold the rule of L5 forwards at position p5 with the rule active(c) -> mark(b). ==> L6 = active^#(f(mark(b),mark(c),c)) -> active^#(f(mark(b),mark(c),c)) [unit] is in U_IR^6. 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 = 6036 popout

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actions

all output return to TRS Standard