Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Logic Programming pair #487096418
details
property
value
status
complete
benchmark
palindrome.pl
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n148.star.cs.uiowa.edu
space
BCGGV05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.78044 seconds
cpu usage
3.91574
user time
3.7542
system time
0.161546
max virtual memory
1.8277284E7
max residence set size
308804.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.pl # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Left Termination of the query pattern palindrome(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) PiDP (9) PiDPToQDPProof [EQUIVALENT, 9 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Clauses: palindrome(Xs) :- reverse(Xs, Xs). reverse(X1s, X2s) :- reverse(X1s, [], X2s). reverse([], Xs, Xs). reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys). Query: palindrome(g) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: palindrome_in_1: (b) reverse_in_2: (b,b) reverse_in_3: (b,b,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs)) reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s)) reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs) reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys)) U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys) U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s) U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs) Pi is empty. Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs)) reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s)) reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs) reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys)) U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys) U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s) U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs) Pi is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: PALINDROME_IN_G(Xs) -> U1_G(Xs, reverse_in_gg(Xs, Xs)) PALINDROME_IN_G(Xs) -> REVERSE_IN_GG(Xs, Xs) REVERSE_IN_GG(X1s, X2s) -> U2_GG(X1s, X2s, reverse_in_ggg(X1s, [], X2s)) REVERSE_IN_GG(X1s, X2s) -> REVERSE_IN_GGG(X1s, [], X2s) REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Logic Programming