details

property	value
status	complete
benchmark	abaaba-aababababaab.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.71938 seconds
cpu usage	4.00589
user time	3.80987
system time	0.196024
max virtual memory	3.8041164E7
max residence set size	318844.0

stage attributes

key	value
starexec-result	YES

output

YES 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4) QTRS (5) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(a(b(a(x1)))))) -> a(a(b(a(b(a(b(a(b(a(a(b(x1)))))))))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(a(b(a(x1)))))) -> b(a(a(b(a(b(a(b(a(b(a(a(x1)))))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(b(a(a(b(a(x1)))))) -> b(a(a(b(a(b(a(b(a(b(a(a(x1)))))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(b(a(a(b(x))))) -> b(a(a(b(a(b(a(b(a(b(a(x))))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(a(b(x))))) -> b(a(a(b(a(b(a(b(a(b(a(x))))))))))) Q is empty. ---------------------------------------- (5) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. The following rules were used to construct the certificate: a(b(a(a(b(x))))) -> b(a(a(b(a(b(a(b(a(b(a(x))))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 544, 545, 546, 547, 548, 549 Node 528 is start node and node 529 is final node. Those nodes are connected through the following edges: * 528 to 530 labelled b_1(0)* 529 to 529 labelled #_1(0)* 530 to 531 labelled a_1(0)* 531 to 532 labelled a_1(0)* 532 to 533 labelled b_1(0)* 533 to 534 labelled a_1(0)* 534 to 535 labelled b_1(0)* 535 to 536 labelled a_1(0)* 536 to 537 labelled b_1(0)* 537 to 538 labelled a_1(0)* 537 to 540 labelled b_1(1)* 538 to 539 labelled b_1(0)* 539 to 529 labelled a_1(0)* 539 to 540 labelled b_1(1)* 540 to 541 labelled a_1(1)* 541 to 542 labelled a_1(1)* 542 to 543 labelled b_1(1)* 543 to 544 labelled a_1(1)* 544 to 545 labelled b_1(1)* 545 to 546 labelled a_1(1)* 546 to 547 labelled b_1(1)* 547 to 548 labelled a_1(1)* 547 to 540 labelled b_1(1)* 548 to 549 labelled b_1(1)* 549 to 529 labelled a_1(1)* 549 to 540 labelled b_1(1) ---------------------------------------- (6) YES popout

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