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	n113.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.07359790802 seconds
cpu usage	5.004760736
max memory	4.66546688E8

stage attributes

key	value
output-size	2361
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) NonTerminationProof [COMPLETE, 0 ms] (2) NO ----------------------------------------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.

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actions

all output return to SRS Standard