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TRS Stand 20472 pair #381712402
details
property
value
status
complete
benchmark
Ex2_Luc02a_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n056.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.379968166351 seconds
cpu usage
0.377436769
max memory
1.3234176E7
stage attributes
key
value
output-size
18881
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o activate : [o] --> o add : [o * o] --> o cons : [o * o] --> o dbl : [o] --> o first : [o * o] --> o n!6220!6220first : [o * o] --> o n!6220!6220terms : [o] --> o nil : [] --> o recip : [o] --> o s : [o] --> o sqr : [o] --> o terms : [o] --> o terms(X) => cons(recip(sqr(X)), n!6220!6220terms(s(X))) sqr(0) => 0 sqr(s(X)) => s(add(sqr(X), dbl(X))) dbl(0) => 0 dbl(s(X)) => s(s(dbl(X))) add(0, X) => X add(s(X), Y) => s(add(X, Y)) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) terms(X) => n!6220!6220terms(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) => terms(X) activate(n!6220!6220first(X, Y)) => first(X, Y) activate(X) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): terms(X) >? cons(recip(sqr(X)), n!6220!6220terms(s(X))) sqr(0) >? 0 sqr(s(X)) >? s(add(sqr(X), dbl(X))) dbl(0) >? 0 dbl(s(X)) >? s(s(dbl(X))) add(0, X) >? X add(s(X), Y) >? s(add(X, Y)) first(0, X) >? nil first(s(X), cons(Y, Z)) >? cons(Y, n!6220!6220first(X, activate(Z))) terms(X) >? n!6220!6220terms(X) first(X, Y) >? n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) >? terms(X) activate(n!6220!6220first(X, Y)) >? first(X, Y) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[nil]] = _|_ [[recip(x_1)]] = x_1 We choose Lex = {} and Mul = {activate, add, cons, dbl, first, n!6220!6220first, n!6220!6220terms, s, sqr, terms}, and the following precedence: activate = first > n!6220!6220first > terms > cons > n!6220!6220terms > sqr > add > dbl > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: terms(X) > cons(sqr(X), n!6220!6220terms(s(X))) sqr(_|_) >= _|_ sqr(s(X)) >= s(add(sqr(X), dbl(X))) dbl(_|_) >= _|_ dbl(s(X)) > s(s(dbl(X))) add(_|_, X) >= X add(s(X), Y) > s(add(X, Y)) first(_|_, X) >= _|_ first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) terms(X) >= n!6220!6220terms(X) first(X, Y) >= n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) > terms(X) activate(n!6220!6220first(X, Y)) > first(X, Y) activate(X) >= X With these choices, we have: 1] terms(X) > cons(sqr(X), n!6220!6220terms(s(X))) because [2], by definition 2] terms*(X) >= cons(sqr(X), n!6220!6220terms(s(X))) because terms > cons, [3] and [6], by (Copy) 3] terms*(X) >= sqr(X) because terms > sqr and [4], by (Copy) 4] terms*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] terms*(X) >= n!6220!6220terms(s(X)) because terms > n!6220!6220terms and [7], by (Copy) 7] terms*(X) >= s(X) because terms > s and [4], by (Copy) 8] sqr(_|_) >= _|_ by (Bot)
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