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TRS Equational pair #487092698
details
property
value
status
complete
benchmark
AC18.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
AProVE_AC_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
5.37144 seconds
cpu usage
11.4656
user time
11.0402
system time
0.425355
max virtual memory
1.834186E7
max residence set size
640900.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ETRS could be proven: (0) ETRS (1) RRRPoloETRSProof [EQUIVALENT, 153 ms] (2) ETRS (3) RRRPoloETRSProof [EQUIVALENT, 42 ms] (4) ETRS (5) RRRPoloETRSProof [EQUIVALENT, 34 ms] (6) ETRS (7) RRRPoloETRSProof [EQUIVALENT, 1047 ms] (8) ETRS (9) EquationalDependencyPairsProof [EQUIVALENT, 100 ms] (10) EDP (11) EUsableRulesReductionPairsProof [EQUIVALENT, 19 ms] (12) EDP (13) ERuleRemovalProof [EQUIVALENT, 17 ms] (14) EDP (15) EDPPoloProof [EQUIVALENT, 27 ms] (16) EDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Equational rewrite system: The TRS R consists of the following rules: 0(S) -> S plus(S, x) -> x plus(0(x), 0(y)) -> 0(plus(x, y)) plus(0(x), 1(y)) -> 1(plus(x, y)) plus(0(x), j(y)) -> j(plus(x, y)) plus(1(x), 1(y)) -> j(plus(1(S), plus(x, y))) plus(j(x), j(y)) -> 1(plus(j(S), plus(x, y))) plus(1(x), j(y)) -> 0(plus(x, y)) opp(S) -> S opp(0(x)) -> 0(opp(x)) opp(1(x)) -> j(opp(x)) opp(j(x)) -> 1(opp(x)) minus(x, y) -> plus(opp(y), x) times(S, x) -> S times(0(x), y) -> 0(times(x, y)) times(1(x), y) -> plus(0(times(x, y)), y) times(j(x), y) -> minus(0(times(x, y)), y) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) ---------------------------------------- (1) RRRPoloETRSProof (EQUIVALENT) The following E TRS is given: Equational rewrite system: The TRS R consists of the following rules: 0(S) -> S plus(S, x) -> x plus(0(x), 0(y)) -> 0(plus(x, y)) plus(0(x), 1(y)) -> 1(plus(x, y)) plus(0(x), j(y)) -> j(plus(x, y)) plus(1(x), 1(y)) -> j(plus(1(S), plus(x, y))) plus(j(x), j(y)) -> 1(plus(j(S), plus(x, y))) plus(1(x), j(y)) -> 0(plus(x, y)) opp(S) -> S opp(0(x)) -> 0(opp(x)) opp(1(x)) -> j(opp(x)) opp(j(x)) -> 1(opp(x)) minus(x, y) -> plus(opp(y), x) times(S, x) -> S times(0(x), y) -> 0(times(x, y)) times(1(x), y) -> plus(0(times(x, y)), y) times(j(x), y) -> minus(0(times(x, y)), y) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly by a polynomial ordering: plus(1(x), j(y)) -> 0(plus(x, y)) opp(1(x)) -> j(opp(x)) opp(j(x)) -> 1(opp(x)) minus(x, y) -> plus(opp(y), x) times(1(x), y) -> plus(0(times(x, y)), y) Used ordering:
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