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TRS Stand 20472 pair #381712739
details
property
value
status
complete
benchmark
secret1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n085.star.cs.uiowa.edu
space
Secret_07_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.423290967941 seconds
cpu usage
0.412937653
max memory
1.5798272E7
stage attributes
key
value
output-size
16817
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 1 : [] --> o 2 : [] --> o D : [o] --> o b : [o * o] --> o c : [o * o] --> o constant : [] --> o div : [o * o] --> o h : [] --> o ln : [o] --> o m : [o * o] --> o opp : [o] --> o pow : [o * o] --> o s : [o] --> o t : [] --> o D(t) => s(h) D(constant) => h D(b(X, Y)) => b(D(X), D(Y)) D(c(X, Y)) => b(c(Y, D(X)), c(X, D(Y))) D(m(X, Y)) => m(D(X), D(Y)) D(opp(X)) => opp(D(X)) D(div(X, Y)) => m(div(D(X), Y), div(c(X, D(Y)), pow(Y, 2))) D(ln(X)) => div(D(X), X) D(pow(X, Y)) => b(c(c(Y, pow(X, m(Y, 1))), D(X)), c(c(pow(X, Y), ln(X)), D(Y))) b(h, X) => X b(X, h) => X b(s(X), s(Y)) => s(s(b(X, Y))) b(b(X, Y), Z) => b(X, b(Y, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): D(t) >? s(h) D(constant) >? h D(b(X, Y)) >? b(D(X), D(Y)) D(c(X, Y)) >? b(c(Y, D(X)), c(X, D(Y))) D(m(X, Y)) >? m(D(X), D(Y)) D(opp(X)) >? opp(D(X)) D(div(X, Y)) >? m(div(D(X), Y), div(c(X, D(Y)), pow(Y, 2))) D(ln(X)) >? div(D(X), X) D(pow(X, Y)) >? b(c(c(Y, pow(X, m(Y, 1))), D(X)), c(c(pow(X, Y), ln(X)), D(Y))) b(h, X) >? X b(X, h) >? X b(s(X), s(Y)) >? s(s(b(X, Y))) b(b(X, Y), Z) >? b(X, b(Y, Z)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[1]] = _|_ [[2]] = _|_ [[h]] = _|_ We choose Lex = {b} and Mul = {D, c, constant, div, ln, m, opp, pow, s, t}, and the following precedence: D = ln = opp = pow > b > c > constant > div > m > t > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: D(t) > s(_|_) D(constant) >= _|_ D(b(X, Y)) > b(D(X), D(Y)) D(c(X, Y)) > b(c(Y, D(X)), c(X, D(Y))) D(m(X, Y)) >= m(D(X), D(Y)) D(opp(X)) >= opp(D(X)) D(div(X, Y)) > m(div(D(X), Y), div(c(X, D(Y)), pow(Y, _|_))) D(ln(X)) >= div(D(X), X) D(pow(X, Y)) >= b(c(c(Y, pow(X, m(Y, _|_))), D(X)), c(c(pow(X, Y), ln(X)), D(Y))) b(_|_, X) >= X b(X, _|_) >= X b(s(X), s(Y)) >= s(s(b(X, Y))) b(b(X, Y), Z) > b(X, b(Y, Z)) With these choices, we have: 1] D(t) > s(_|_) because [2], by definition 2] D*(t) >= s(_|_) because [3], by (Select) 3] t >= s(_|_) because [4], by (Star) 4] t* >= s(_|_) because t > s and [5], by (Copy) 5] t* >= _|_ by (Bot) 6] D(constant) >= _|_ by (Bot) 7] D(b(X, Y)) > b(D(X), D(Y)) because [8], by definition 8] D*(b(X, Y)) >= b(D(X), D(Y)) because D > b, [9] and [13], by (Copy) 9] D*(b(X, Y)) >= D(X) because D in Mul and [10], by (Stat) 10] b(X, Y) > X because [11], by definition
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