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TRS Stand 20472 pair #381715377
details
property
value
status
complete
benchmark
Ex4_7_77_Bor03_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n077.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.68622708321 seconds
cpu usage
3.476288303
max memory
1.87867136E8
stage attributes
key
value
output-size
4663
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSToCSRProof [EQUIVALENT, 0 ms] (2) CSR (3) CSRRRRProof [EQUIVALENT, 61 ms] (4) CSR (5) CSRRRRProof [EQUIVALENT, 0 ms] (6) CSR (7) RisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(zeros) -> mark(cons(0, zeros)) active(tail(cons(X, XS))) -> mark(XS) active(cons(X1, X2)) -> cons(active(X1), X2) active(tail(X)) -> tail(active(X)) cons(mark(X1), X2) -> mark(cons(X1, X2)) tail(mark(X)) -> mark(tail(X)) proper(zeros) -> ok(zeros) proper(cons(X1, X2)) -> cons(proper(X1), proper(X2)) proper(0) -> ok(0) proper(tail(X)) -> tail(proper(X)) cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) tail(ok(X)) -> ok(tail(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. ---------------------------------------- (1) QTRSToCSRProof (EQUIVALENT) The following Q TRS is given: Q restricted rewrite system: The TRS R consists of the following rules: active(zeros) -> mark(cons(0, zeros)) active(tail(cons(X, XS))) -> mark(XS) active(cons(X1, X2)) -> cons(active(X1), X2) active(tail(X)) -> tail(active(X)) cons(mark(X1), X2) -> mark(cons(X1, X2)) tail(mark(X)) -> mark(tail(X)) proper(zeros) -> ok(zeros) proper(cons(X1, X2)) -> cons(proper(X1), proper(X2)) proper(0) -> ok(0) proper(tail(X)) -> tail(proper(X)) cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) tail(ok(X)) -> ok(tail(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. Special symbols used for the transformation (see [GM04]): top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1 The replacement map contains the following entries: zeros: empty set cons: {1} 0: empty set tail: {1} The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound). ---------------------------------------- (2) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: zeros -> cons(0, zeros) tail(cons(X, XS)) -> XS The replacement map contains the following entries: zeros: empty set cons: {1} 0: empty set tail: {1} ---------------------------------------- (3) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system:
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