details

property	value
status	complete
benchmark	z124.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	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.9028 seconds
cpu usage	3.79396
user time	3.60922
system time	0.184742
max virtual memory	1.8810352E7
max residence set size	270412.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: q0(a(x1)) -> x(q1(x1)) q1(a(x1)) -> a(q1(x1)) q1(y(x1)) -> y(q1(x1)) a(q1(b(x1))) -> q2(a(y(x1))) a(q2(a(x1))) -> q2(a(a(x1))) a(q2(y(x1))) -> q2(a(y(x1))) y(q1(b(x1))) -> q2(y(y(x1))) y(q2(a(x1))) -> q2(y(a(x1))) y(q2(y(x1))) -> q2(y(y(x1))) q2(x(x1)) -> x(q0(x1)) q0(y(x1)) -> y(q3(x1)) q3(y(x1)) -> y(q3(x1)) q3(bl(x1)) -> bl(q4(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(q0(x1)) -> q1(x(x1)) a(q1(x1)) -> q1(a(x1)) y(q1(x1)) -> q1(y(x1)) b(q1(a(x1))) -> y(a(q2(x1))) a(q2(a(x1))) -> a(a(q2(x1))) y(q2(a(x1))) -> y(a(q2(x1))) b(q1(y(x1))) -> y(y(q2(x1))) a(q2(y(x1))) -> a(y(q2(x1))) y(q2(y(x1))) -> y(y(q2(x1))) x(q2(x1)) -> q0(x(x1)) y(q0(x1)) -> q3(y(x1)) y(q3(x1)) -> q3(y(x1)) bl(q3(x1)) -> q4(bl(x1)) Q is empty. ---------------------------------------- (3) 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(q0(x1)) -> q1(x(x1)) a(q1(x1)) -> q1(a(x1)) y(q1(x1)) -> q1(y(x1)) b(q1(a(x1))) -> y(a(q2(x1))) a(q2(a(x1))) -> a(a(q2(x1))) y(q2(a(x1))) -> y(a(q2(x1))) b(q1(y(x1))) -> y(y(q2(x1))) a(q2(y(x1))) -> a(y(q2(x1))) y(q2(y(x1))) -> y(y(q2(x1))) x(q2(x1)) -> q0(x(x1)) y(q0(x1)) -> q3(y(x1)) y(q3(x1)) -> q3(y(x1)) bl(q3(x1)) -> q4(bl(x1)) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 Node 2 is start node and node 3 is final node. Those nodes are connected through the following edges: * 2 to 4 labelled q1_1(0), q0_1(0), q3_1(0)* 2 to 5 labelled y_1(0), a_1(0)* 2 to 7 labelled y_1(0), a_1(0)* 2 to 9 labelled q4_1(0)* 3 to 3 labelled #_1(0)* 4 to 3 labelled x_1(0), a_1(0), y_1(0)* 4 to 10 labelled q0_1(1), q1_1(1), q3_1(1)* 4 to 11 labelled a_1(1), y_1(1)* 4 to 13 labelled a_1(1), y_1(1)* 5 to 6 labelled a_1(0)* 5 to 11 labelled a_1(1)* 5 to 13 labelled a_1(1)* 6 to 3 labelled q2_1(0)* 7 to 8 labelled y_1(0)* 7 to 11 labelled y_1(1)* 7 to 13 labelled y_1(1)* 8 to 3 labelled q2_1(0)* 9 to 3 labelled bl_1(0)* 9 to 15 labelled q4_1(1)* 10 to 3 labelled x_1(1), a_1(1), y_1(1)* 10 to 10 labelled q0_1(1), q1_1(1), q3_1(1)* 10 to 11 labelled a_1(1), y_1(1)* 10 to 13 labelled a_1(1), y_1(1)* 11 to 12 labelled a_1(1)* 11 to 11 labelled a_1(1)* 11 to 13 labelled a_1(1)* 12 to 3 labelled q2_1(1)* 13 to 14 labelled y_1(1)* 13 to 11 labelled y_1(1)* 13 to 13 labelled y_1(1)* 14 to 3 labelled q2_1(1)* 15 to 3 labelled bl_1(1)* 15 to 15 labelled q4_1(1) ---------------------------------------- (4) YES popout

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