details

property	value
status	complete
benchmark	caabccaabc-aabccaabcaabcc.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.06683516502 seconds
cpu usage	5.504508323
max memory	5.0198528E8

stage attributes

key	value
output-size	2968
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Strip Symbols Proof [SOUND, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]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: c(a(a(b(c(c(a(a(b(c(x1)))))))))) -> a(a(b(c(c(a(a(b(c(a(a(b(c(c(x1)))))))))))))) Q is empty. ---------------------------------------- (1) Strip Symbols Proof (SOUND) We were given the following TRS: c(a(a(b(c(c(a(a(b(c(x1)))))))))) -> a(a(b(c(c(a(a(b(c(a(a(b(c(c(x1)))))))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: c(a(a(b(c(c(a(a(b(x))))))))) -> a(a(b(c(c(a(a(b(c(a(a(b(c(x))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(a(a(b(c(c(a(a(b(x))))))))) -> a(a(b(c(c(a(a(b(c(a(a(b(c(x))))))))))))) 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: c(a(a(b(c(c(a(a(b(x))))))))) -> a(a(b(c(c(a(a(b(c(a(a(b(c(x))))))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 3, 4, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132 Node 3 is start node and node 4 is final node. Those nodes are connected through the following edges: * 3 to 109 labelled a_1(0)* 4 to 4 labelled #_1(0)* 109 to 110 labelled a_1(0)* 110 to 111 labelled b_1(0)* 111 to 112 labelled c_1(0)* 112 to 113 labelled c_1(0)* 113 to 114 labelled a_1(0)* 114 to 115 labelled a_1(0)* 115 to 116 labelled b_1(0)* 116 to 117 labelled c_1(0)* 116 to 121 labelled a_1(1)* 117 to 118 labelled a_1(0)* 118 to 119 labelled a_1(0)* 119 to 120 labelled b_1(0)* 120 to 4 labelled c_1(0)* 120 to 121 labelled a_1(1)* 121 to 122 labelled a_1(1)* 122 to 123 labelled b_1(1)* 123 to 124 labelled c_1(1)* 124 to 125 labelled c_1(1)* 125 to 126 labelled a_1(1)* 126 to 127 labelled a_1(1)* 127 to 128 labelled b_1(1)* 128 to 129 labelled c_1(1)* 128 to 121 labelled a_1(1)* 129 to 130 labelled a_1(1)* 130 to 131 labelled a_1(1)* 131 to 132 labelled b_1(1)* 132 to 4 labelled c_1(1)* 132 to 121 labelled a_1(1) ---------------------------------------- (4) YES popout

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