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TRS Standard pair #487068413
details
property
value
status
complete
benchmark
recursion-5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n183.star.cs.uiowa.edu
space
TCT_12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.8622 seconds
cpu usage
3.88385
user time
3.72672
system time
0.157131
max virtual memory
1.8408912E7
max residence set size
247544.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 122 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f_0(x) -> a f_1(x) -> g_1(x, x) g_1(s(x), y) -> b(f_0(y), g_1(x, y)) f_2(x) -> g_2(x, x) g_2(s(x), y) -> b(f_1(y), g_2(x, y)) f_3(x) -> g_3(x, x) g_3(s(x), y) -> b(f_2(y), g_3(x, y)) f_4(x) -> g_4(x, x) g_4(s(x), y) -> b(f_3(y), g_4(x, y)) f_5(x) -> g_5(x, x) g_5(s(x), y) -> b(f_4(y), g_5(x, y)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Quasi precedence: f_5_1 > g_5_2 > f_4_1 > [f_3_1, g_4_2] > [f_2_1, g_3_2] > [f_1_1, g_2_2] > [f_0_1, a, g_1_2] > b_2 Status: f_0_1: multiset status a: multiset status f_1_1: multiset status g_1_2: multiset status s_1: multiset status b_2: multiset status f_2_1: multiset status g_2_2: multiset status f_3_1: multiset status g_3_2: multiset status f_4_1: [1] g_4_2: multiset status f_5_1: [1] g_5_2: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f_0(x) -> a f_1(x) -> g_1(x, x) g_1(s(x), y) -> b(f_0(y), g_1(x, y)) f_2(x) -> g_2(x, x) g_2(s(x), y) -> b(f_1(y), g_2(x, y)) f_3(x) -> g_3(x, x) g_3(s(x), y) -> b(f_2(y), g_3(x, y)) f_4(x) -> g_4(x, x) g_4(s(x), y) -> b(f_3(y), g_4(x, y)) f_5(x) -> g_5(x, x) g_5(s(x), y) -> b(f_4(y), g_5(x, y)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (4) YES
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