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TRS Inner 89993 pair #381733558
details
property
value
status
complete
benchmark
#4.35.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n037.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.379144907 seconds
cpu usage
5.931659769
max memory
3.52366592E8
stage attributes
key
value
output-size
16956
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 71 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: and(true, y) -> y and(false, y) -> false eq(nil, nil) -> true eq(cons(t, l), nil) -> false eq(nil, cons(t, l)) -> false eq(cons(t, l), cons(t', l')) -> and(eq(t, t'), eq(l, l')) eq(var(l), var(l')) -> eq(l, l') eq(var(l), apply(t, s)) -> false eq(var(l), lambda(x, t)) -> false eq(apply(t, s), var(l)) -> false eq(apply(t, s), apply(t', s')) -> and(eq(t, t'), eq(s, s')) eq(apply(t, s), lambda(x, t)) -> false eq(lambda(x, t), var(l)) -> false eq(lambda(x, t), apply(t, s)) -> false eq(lambda(x, t), lambda(x', t')) -> and(eq(x, x'), eq(t, t')) if(true, var(k), var(l')) -> var(k) if(false, var(k), var(l')) -> var(l') ren(var(l), var(k), var(l')) -> if(eq(l, l'), var(k), var(l')) ren(x, y, apply(t, s)) -> apply(ren(x, y, t), ren(x, y, s)) ren(x, y, lambda(z, t)) -> lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t))) The set Q consists of the following terms: and(true, x0) and(false, x0) eq(nil, nil) eq(cons(x0, x1), nil) eq(nil, cons(x0, x1)) eq(cons(x0, x1), cons(x2, x3)) eq(var(x0), var(x1)) eq(var(x0), apply(x1, x2)) eq(var(x0), lambda(x1, x2)) eq(apply(x0, x1), var(x2)) eq(apply(x0, x1), apply(x2, x3)) eq(apply(x0, x1), lambda(x2, x0)) eq(lambda(x0, x1), var(x2)) eq(lambda(x0, x1), apply(x1, x2)) eq(lambda(x0, x1), lambda(x2, x3)) if(true, var(x0), var(x1)) if(false, var(x0), var(x1)) ren(var(x0), var(x1), var(x2)) ren(x0, x1, apply(x2, x3)) ren(x0, x1, lambda(x2, x3)) ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: EQ(cons(t, l), cons(t', l')) -> AND(eq(t, t'), eq(l, l')) EQ(cons(t, l), cons(t', l')) -> EQ(t, t') EQ(cons(t, l), cons(t', l')) -> EQ(l, l') EQ(var(l), var(l')) -> EQ(l, l') EQ(apply(t, s), apply(t', s')) -> AND(eq(t, t'), eq(s, s'))
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