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TRS Outer 67415 pair #381917995
details
property
value
status
complete
benchmark
ex8.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n036.star.cs.uiowa.edu
space
Zantema_08
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.4680120945 seconds
cpu usage
3.875350204
max memory
2.63077888E8
stage attributes
key
value
output-size
12116
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could be proven: (0) OTRS (1) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 3 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) RFCMatchBoundsDPProof [EQUIVALENT, 1 ms] (12) YES ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: f(g(x)) -> g(g(x)) g(x) -> f(f(x)) f(f(x)) -> x Outermost Strategy. ---------------------------------------- (1) Thiemann-SpecialC-Transformation (EQUIVALENT) We applied the Thiemann-SpecialC transformation to transform the outermost TRS to an innermost TRS. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1)) -> check_f(redex_f(x_1)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_f(g(x)) -> result_f(g(g(x))) redex_g(x) -> result_g(f(f(x))) redex_f(f(x)) -> result_f(x) check_f(result_f(x)) -> go_up(x) check_g(result_g(x)) -> go_up(x) check_f(redex_f(x_1)) -> in_f_1(reduce(x_1)) check_g(redex_g(x_1)) -> in_g_1(reduce(x_1)) in_f_1(go_up(x_1)) -> go_up(f(x_1)) in_g_1(go_up(x_1)) -> go_up(g(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0)) reduce(g(x0)) redex_f(g(x0)) redex_g(x0) redex_f(f(x0)) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0)) in_f_1(go_up(x0)) in_g_1(go_up(x0)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) TOP(go_up(x)) -> REDUCE(x) REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1)) REDUCE(f(x_1)) -> REDEX_F(x_1) REDUCE(g(x_1)) -> CHECK_G(redex_g(x_1)) REDUCE(g(x_1)) -> REDEX_G(x_1) CHECK_F(redex_f(x_1)) -> IN_F_1(reduce(x_1))
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