details

property	value
status	complete
benchmark	Ex2_Luc03b_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n049.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.122072935104 seconds
cpu usage	0.116731282
max memory	2990080.0

stage attributes

key	value
output-size	10121
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 from : [o] --> o fst : [o * o] --> o len : [o] --> o n!6220!6220add : [o * o] --> o n!6220!6220from : [o] --> o n!6220!6220fst : [o * o] --> o n!6220!6220len : [o] --> o nil : [] --> o s : [o] --> o fst(0, X) => nil fst(s(X), cons(Y, Z)) => cons(Y, n!6220!6220fst(activate(X), activate(Z))) from(X) => cons(X, n!6220!6220from(s(X))) add(0, X) => X add(s(X), Y) => s(n!6220!6220add(activate(X), Y)) len(nil) => 0 len(cons(X, Y)) => s(n!6220!6220len(activate(Y))) fst(X, Y) => n!6220!6220fst(X, Y) from(X) => n!6220!6220from(X) add(X, Y) => n!6220!6220add(X, Y) len(X) => n!6220!6220len(X) activate(n!6220!6220fst(X, Y)) => fst(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220add(X, Y)) => add(X, Y) activate(n!6220!6220len(X)) => len(X) 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]): fst(0, X) >? nil fst(s(X), cons(Y, Z)) >? cons(Y, n!6220!6220fst(activate(X), activate(Z))) from(X) >? cons(X, n!6220!6220from(s(X))) add(0, X) >? X add(s(X), Y) >? s(n!6220!6220add(activate(X), Y)) len(nil) >? 0 len(cons(X, Y)) >? s(n!6220!6220len(activate(Y))) fst(X, Y) >? n!6220!6220fst(X, Y) from(X) >? n!6220!6220from(X) add(X, Y) >? n!6220!6220add(X, Y) len(X) >? n!6220!6220len(X) activate(n!6220!6220fst(X, Y)) >? fst(X, Y) activate(n!6220!6220from(X)) >? from(X) activate(n!6220!6220add(X, Y)) >? add(X, Y) activate(n!6220!6220len(X)) >? len(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 2 activate = \y0.2y0 add = \y0y1.2y0 + 2y1 cons = \y0y1.y0 + y1 from = \y0.2y0 fst = \y0y1.2 + 2y0 + 2y1 len = \y0.2y0 n!6220!6220add = \y0y1.y0 + 2y1 n!6220!6220from = \y0.y0 n!6220!6220fst = \y0y1.1 + y0 + y1 n!6220!6220len = \y0.y0 nil = 2 s = \y0.y0 Using this interpretation, the requirements translate to: [[fst(0, _x0)]] = 6 + 2x0 > 2 = [[nil]] [[fst(s(_x0), cons(_x1, _x2))]] = 2 + 2x0 + 2x1 + 2x2 > 1 + x1 + 2x0 + 2x2 = [[cons(_x1, n!6220!6220fst(activate(_x0), activate(_x2)))]] [[from(_x0)]] = 2x0 >= 2x0 = [[cons(_x0, n!6220!6220from(s(_x0)))]] [[add(0, _x0)]] = 4 + 2x0 > x0 = [[_x0]] [[add(s(_x0), _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[s(n!6220!6220add(activate(_x0), _x1))]] [[len(nil)]] = 4 > 2 = [[0]] [[len(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[s(n!6220!6220len(activate(_x1)))]] [[fst(_x0, _x1)]] = 2 + 2x0 + 2x1 > 1 + x0 + x1 = [[n!6220!6220fst(_x0, _x1)]] [[from(_x0)]] = 2x0 >= x0 = [[n!6220!6220from(_x0)]] [[add(_x0, _x1)]] = 2x0 + 2x1 >= x0 + 2x1 = [[n!6220!6220add(_x0, _x1)]] [[len(_x0)]] = 2x0 >= x0 = [[n!6220!6220len(_x0)]] [[activate(n!6220!6220fst(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[fst(_x0, _x1)]] [[activate(n!6220!6220from(_x0))]] = 2x0 >= 2x0 = [[from(_x0)]] [[activate(n!6220!6220add(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 2x1 = [[add(_x0, _x1)]] [[activate(n!6220!6220len(_x0))]] = 2x0 >= 2x0 = [[len(_x0)]] [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] popout

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