details

property	value
status	complete
benchmark	turing_mult.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.67209 seconds
cpu usage	3.57051
user time	3.38926
system time	0.181255
max virtual memory	1.8676764E7
max residence set size	246612.0

stage attributes

key	value
starexec-result	NO

output

NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) NonTerminationProof [COMPLETE, 0 ms] (2) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(q0(0(x1))) -> 0(0(q0(x1))) 0(q0(1(x1))) -> 0(1(q0(x1))) 1(q0(0(x1))) -> 0(0(q1(x1))) 1(q0(1(x1))) -> 0(1(q1(x1))) 1(q1(0(x1))) -> 1(0(q1(x1))) 1(q1(1(x1))) -> 1(1(q1(x1))) 0(q1(0(x1))) -> 0(0(q2(x1))) 0(q1(1(x1))) -> 0(1(q2(x1))) 1(q2(0(x1))) -> 1(0(q2(x1))) 1(q2(1(x1))) -> 1(1(q2(x1))) 0(q2(x1)) -> q3(1(x1)) 1(q3(x1)) -> q3(1(x1)) 0(q3(x1)) -> q4(0(x1)) 1(q4(x1)) -> q4(1(x1)) 0(q4(0(x1))) -> 1(0(q5(x1))) 0(q4(1(x1))) -> 1(1(q5(x1))) 1(q5(0(x1))) -> 0(0(q1(x1))) 1(q5(1(x1))) -> 0(1(q1(x1))) 0(q5(x1)) -> q6(0(x1)) 1(q6(x1)) -> q6(1(x1)) 1(q7(0(x1))) -> 0(0(q8(x1))) 1(q7(1(x1))) -> 0(1(q8(x1))) 0(q8(x1)) -> 0(q0(x1)) 1(q8(0(x1))) -> 1(0(q8(x1))) 1(q8(1(x1))) -> 1(1(q8(x1))) 0(q6(x1)) -> q9(0(x1)) 0(q9(0(x1))) -> 1(0(q7(x1))) 0(q9(1(x1))) -> 1(1(q7(x1))) 1(q9(x1)) -> q9(1(x1)) h(q0(x1)) -> h(0(q0(x1))) q0(h(x1)) -> q0(0(h(x1))) h(q1(x1)) -> h(0(q1(x1))) q1(h(x1)) -> q1(0(h(x1))) h(q2(x1)) -> h(0(q2(x1))) q2(h(x1)) -> q2(0(h(x1))) h(q3(x1)) -> h(0(q3(x1))) q3(h(x1)) -> q3(0(h(x1))) h(q4(x1)) -> h(0(q4(x1))) q4(h(x1)) -> q4(0(h(x1))) h(q5(x1)) -> h(0(q5(x1))) q5(h(x1)) -> q5(0(h(x1))) h(q6(x1)) -> h(0(q6(x1))) q6(h(x1)) -> q6(0(h(x1))) Q is empty. ---------------------------------------- (1) NonTerminationProof (COMPLETE) We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite. Found the self-embedding DerivationStructure: "0 q0 0 h -> 0 0 q0 0 h" 0 q0 0 h -> 0 0 q0 0 h by OverlapClosure OC 2"0 q0 0 -> 0 0 q0 by original rule (OC 1)""q0 h -> q0 0 h by original rule (OC 1)" ---------------------------------------- (2) NO popout

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

job log

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actions

all output return to SRS Standard