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TRS Conte Sensi 17651 pair #381733361
details
property
value
status
complete
benchmark
Ex49_GM04.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.49526596069 seconds
cpu usage
3.648451439
max memory
2.27786752E8
stage attributes
key
value
output-size
7634
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given CSR could be proven: (0) CSR (1) CSRRRRProof [EQUIVALENT, 54 ms] (2) CSR (3) CSRRRRProof [EQUIVALENT, 0 ms] (4) CSR (5) CSRRRRProof [EQUIVALENT, 3 ms] (6) CSR (7) CSRRRRProof [EQUIVALENT, 0 ms] (8) CSR (9) CSRRRRProof [EQUIVALENT, 2 ms] (10) CSR (11) RisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: minus(0, Y) -> 0 minus(s(X), s(Y)) -> minus(X, Y) geq(X, 0) -> true geq(0, s(Y)) -> false geq(s(X), s(Y)) -> geq(X, Y) div(0, s(Y)) -> 0 div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) if(true, X, Y) -> X if(false, X, Y) -> Y The replacement map contains the following entries: minus: empty set 0: empty set s: {1} geq: empty set true: empty set false: empty set div: {1} if: {1} ---------------------------------------- (1) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system: The TRS R consists of the following rules: minus(0, Y) -> 0 minus(s(X), s(Y)) -> minus(X, Y) geq(X, 0) -> true geq(0, s(Y)) -> false geq(s(X), s(Y)) -> geq(X, Y) div(0, s(Y)) -> 0 div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) if(true, X, Y) -> X if(false, X, Y) -> Y The replacement map contains the following entries: minus: empty set 0: empty set s: {1} geq: empty set true: empty set false: empty set div: {1} if: {1} Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(div(x_1, x_2)) = x_1 + x_2 POL(false) = 0 POL(geq(x_1, x_2)) = 0 POL(if(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(minus(x_1, x_2)) = x_1 POL(s(x_1)) = 1 + x_1 POL(true) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: minus(s(X), s(Y)) -> minus(X, Y) div(0, s(Y)) -> 0
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