details

property	value
status	complete
benchmark	ExIntrod_GM04_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	1.54110383987 seconds
cpu usage	5.030751153
max memory	6.95185408E8

stage attributes

key	value
output-size	5038
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] adx(cons(_0,n__zeros)) -> adx(cons(n__0,n__zeros)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->n__0}. We have r|p = adx(cons(n__0,n__zeros)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = adx(cons(_0,n__zeros)) 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... ## 1 initial DP problem to solve. ## First, we try to decompose this problem into smaller problems. ## Round 1 [1 DP problem]: ## DP problem: Dependency pairs = [adx^#(cons(_0,_1)) -> incr^#(cons(activate(_0),n__adx(activate(_1)))), activate^#(n__adx(_0)) -> adx^#(activate(_0)), incr^#(cons(_0,_1)) -> activate^#(_0), activate^#(n__incr(_0)) -> incr^#(activate(_0)), activate^#(n__incr(_0)) -> activate^#(_0), activate^#(n__adx(_0)) -> activate^#(_0), adx^#(cons(_0,_1)) -> activate^#(_1), adx^#(cons(_0,_1)) -> activate^#(_0), incr^#(cons(_0,_1)) -> activate^#(_1)] TRS = {nats -> adx(zeros), zeros -> cons(n__0,n__zeros), incr(cons(_0,_1)) -> cons(n__s(activate(_0)),n__incr(activate(_1))), adx(cons(_0,_1)) -> incr(cons(activate(_0),n__adx(activate(_1)))), hd(cons(_0,_1)) -> activate(_0), tl(cons(_0,_1)) -> activate(_1), 0 -> n__0, zeros -> n__zeros, s(_0) -> n__s(_0), incr(_0) -> n__incr(_0), adx(_0) -> n__adx(_0), activate(n__0) -> 0, activate(n__zeros) -> zeros, activate(n__s(_0)) -> s(_0), activate(n__incr(_0)) -> incr(activate(_0)), activate(n__adx(_0)) -> adx(activate(_0)), activate(_0) -> _0} ## 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... This DP problem is too complex! Aborting! 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=false, max=-1) # max_depth=4, unfold_variables=false: # Iteration 0: no loop found, 9 unfolded rules generated. # Iteration 1: no loop found, 53 unfolded rules generated. # Iteration 2: no loop found, 162 unfolded rules generated. # Iteration 3: no loop found, 647 unfolded rules generated. # Iteration 4: no loop found, 1756 unfolded rules generated. # Iteration 5: no loop found, 5357 unfolded rules generated. # Iteration 6: success, found a loop, 13395 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = adx^#(cons(_0,_1)) -> incr^#(cons(activate(_0),n__adx(activate(_1)))) [trans] is in U_IR^0. D = incr^#(cons(_0,_1)) -> activate^#(_1) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = adx^#(cons(_0,_1)) -> activate^#(n__adx(activate(_1))) [trans] is in U_IR^1. D = activate^#(n__adx(_0)) -> adx^#(activate(_0)) is a dependency pair of IR. We build a composed triple from L1 and D. ==> L2 = [adx^#(cons(_0,_1)) -> activate^#(n__adx(activate(_1))), activate^#(n__adx(_2)) -> adx^#(activate(_2))] [comp] is in U_IR^2. Let p2 = [0, 0]. We unfold the first rule of L2 forwards at position p2 with the rule activate(n__zeros) -> zeros. ==> L3 = [adx^#(cons(_0,n__zeros)) -> activate^#(n__adx(zeros)), activate^#(n__adx(_1)) -> adx^#(activate(_1))] [comp] is in U_IR^3. Let p3 = [0, 0]. We unfold the first rule of L3 forwards at position p3 with the rule zeros -> cons(n__0,n__zeros). ==> L4 = adx^#(cons(_0,n__zeros)) -> adx^#(activate(cons(n__0,n__zeros))) [trans] is in U_IR^4. D = adx^#(cons(_0,_1)) -> activate^#(_1) is a dependency pair of IR. We build a composed triple from L4 and D. ==> L5 = [adx^#(cons(_0,n__zeros)) -> adx^#(activate(cons(n__0,n__zeros))), adx^#(cons(_1,_2)) -> activate^#(_2)] [comp] is in U_IR^5. Let p5 = [0]. We unfold the first rule of L5 forwards at position p5 with the rule activate(_0) -> _0. ==> L6 = [adx^#(cons(_0,n__zeros)) -> adx^#(cons(n__0,n__zeros)), adx^#(cons(_1,_2)) -> activate^#(_2)] [comp] 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 = 45853 popout

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actions

all output return to TRS Standard