details

property	value
status	complete
benchmark	sat.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n182.star.cs.uiowa.edu
space	TCT_12

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	0.190097 seconds
cpu usage	0.272458
user time	0.234918
system time	0.03754
max virtual memory	113188.0
max residence set size	59260.0

stage attributes

key	value
starexec-result	YES

output

YES Prover = TRS(tech=PATTERN_RULES, nb_unfoldings=unlimited, max_nb_unfolded_rules=200) ** BEGIN proof argument ** All the DP problems were proved finite. As all the involved DP processors are sound, the TRS under analysis terminates. ** END proof argument ** ** BEGIN proof description ** ## Searching for a generalized rewrite rule (a rule whose right-hand side contains a variable that does not occur in the left-hand side)... No generalized rewrite rule found! ## Applying the DP framework... ## Round 1: ## DP problem: Dependency pairs = [verify^#(cons(_0,_1)) -> verify^#(_1)] TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [guess^#(cons(_0,_1)) -> guess^#(_1)] TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [choice^#(cons(_0,_1)) -> choice^#(_1)] TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [member^#(_0,cons(_1,_2)) -> member^#(_0,_2)] TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [eq^#(O(_0),0(_1)) -> eq^#(_0,_1), eq^#(1(_0),1(_1)) -> eq^#(_0,_1)] TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ** END proof description ** Proof stopped at iteration 0 Number of unfolded rules generated by this proof = 0 Number of unfolded rules generated by all the parallel proofs = 0 popout

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all output return to TRS Standard