details

property	value
status	complete
benchmark	polycounter-10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	TCT_12

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	1.20827388763 seconds
cpu usage	1.204470368
max memory	6.2005248E7

stage attributes

key	value
output-size	14809
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 f : [o * o * o * o * o * o * o * o * o * o] --> o s : [o] --> o f(s(X), Y, Z, U, V, W, Q, R, S, T) => f(X, Y, Z, U, V, W, Q, R, S, T) f(0, s(X), Y, Z, U, V, W, Q, R, S) => f(X, X, Y, Z, U, V, W, Q, R, S) f(0, 0, s(X), Y, Z, U, V, W, Q, R) => f(X, X, X, Y, Z, U, V, W, Q, R) f(0, 0, 0, s(X), Y, Z, U, V, W, Q) => f(X, X, X, X, Y, Z, U, V, W, Q) f(0, 0, 0, 0, s(X), Y, Z, U, V, W) => f(X, X, X, X, X, Y, Z, U, V, W) f(0, 0, 0, 0, 0, s(X), Y, Z, U, V) => f(X, X, X, X, X, X, Y, Z, U, V) f(0, 0, 0, 0, 0, 0, s(X), Y, Z, U) => f(X, X, X, X, X, X, X, Y, Z, U) f(0, 0, 0, 0, 0, 0, 0, s(X), Y, Z) => f(X, X, X, X, X, X, X, X, Y, Z) f(0, 0, 0, 0, 0, 0, 0, 0, s(X), Y) => f(X, X, X, X, X, X, X, X, X, Y) f(0, 0, 0, 0, 0, 0, 0, 0, 0, s(X)) => f(X, X, X, X, X, X, X, X, X, X) f(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) => 0 We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(s(X), Y, Z, U, V, W, Q, R, S, T) >? f(X, Y, Z, U, V, W, Q, R, S, T) f(0, s(X), Y, Z, U, V, W, Q, R, S) >? f(X, X, Y, Z, U, V, W, Q, R, S) f(0, 0, s(X), Y, Z, U, V, W, Q, R) >? f(X, X, X, Y, Z, U, V, W, Q, R) f(0, 0, 0, s(X), Y, Z, U, V, W, Q) >? f(X, X, X, X, Y, Z, U, V, W, Q) f(0, 0, 0, 0, s(X), Y, Z, U, V, W) >? f(X, X, X, X, X, Y, Z, U, V, W) f(0, 0, 0, 0, 0, s(X), Y, Z, U, V) >? f(X, X, X, X, X, X, Y, Z, U, V) f(0, 0, 0, 0, 0, 0, s(X), Y, Z, U) >? f(X, X, X, X, X, X, X, Y, Z, U) f(0, 0, 0, 0, 0, 0, 0, s(X), Y, Z) >? f(X, X, X, X, X, X, X, X, Y, Z) f(0, 0, 0, 0, 0, 0, 0, 0, s(X), Y) >? f(X, X, X, X, X, X, X, X, X, Y) f(0, 0, 0, 0, 0, 0, 0, 0, 0, s(X)) >? f(X, X, X, X, X, X, X, X, X, X) f(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) >? 0 about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {f, s}, and the following precedence: s > f Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: f(s(X), Y, Z, U, V, W, Q, R, S, T) >= f(X, Y, Z, U, V, W, Q, R, S, T) f(_|_, s(X), Y, Z, U, V, W, Q, R, S) > f(X, X, Y, Z, U, V, W, Q, R, S) f(_|_, _|_, s(X), Y, Z, U, V, W, Q, R) > f(X, X, X, Y, Z, U, V, W, Q, R) f(_|_, _|_, _|_, s(X), Y, Z, U, V, W, Q) > f(X, X, X, X, Y, Z, U, V, W, Q) f(_|_, _|_, _|_, _|_, s(X), Y, Z, U, V, W) > f(X, X, X, X, X, Y, Z, U, V, W) f(_|_, _|_, _|_, _|_, _|_, s(X), Y, Z, U, V) >= f(X, X, X, X, X, X, Y, Z, U, V) f(_|_, _|_, _|_, _|_, _|_, _|_, s(X), Y, Z, U) >= f(X, X, X, X, X, X, X, Y, Z, U) f(_|_, _|_, _|_, _|_, _|_, _|_, _|_, s(X), Y, Z) > f(X, X, X, X, X, X, X, X, Y, Z) f(_|_, _|_, _|_, _|_, _|_, _|_, _|_, _|_, s(X), Y) > f(X, X, X, X, X, X, X, X, X, Y) f(_|_, _|_, _|_, _|_, _|_, _|_, _|_, _|_, _|_, s(X)) >= f(X, X, X, X, X, X, X, X, X, X) f(_|_, _|_, _|_, _|_, _|_, _|_, _|_, _|_, _|_, _|_) >= _|_ With these choices, we have: 1] f(s(X), Y, Z, U, V, W, Q, R, S, T) >= f(X, Y, Z, U, V, W, Q, R, S, T) because f in Mul, [2], [5], [6], [7], [8], [9], [10], [11], [12] and [13], by (Fun) 2] s(X) >= X because [3], by (Star) 3] s*(X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] Y >= Y by (Meta) 6] Z >= Z by (Meta) 7] U >= U by (Meta) 8] V >= V by (Meta) 9] W >= W by (Meta) 10] Q >= Q by (Meta) 11] R >= R by (Meta) 12] S >= S by (Meta) 13] T >= T by (Meta) 14] f(_|_, s(X), Y, Z, U, V, W, Q, R, S) > f(X, X, Y, Z, U, V, W, Q, R, S) because [15], by definition 15] f*(_|_, s(X), Y, Z, U, V, W, Q, R, S) >= f(X, X, Y, Z, U, V, W, Q, R, S) because f in Mul, [16], [16], [6], [7], [8], [9], [10], [11], [12] and [13], by (Stat) 16] s(X) > X because [17], by definition 17] s*(X) >= X because [5], by (Select) 18] f(_|_, _|_, s(X), Y, Z, U, V, W, Q, R) > f(X, X, X, Y, Z, U, V, W, Q, R) because [19], by definition 19] f*(_|_, _|_, s(X), Y, Z, U, V, W, Q, R) >= f(X, X, X, Y, Z, U, V, W, Q, R) because f in Mul, [20], [20], [20], [7], [8], [9], [10], [11], [12] and [13], by (Stat) 20] s(X) > X because [21], by definition 21] s*(X) >= X because [6], by (Select) 22] f(_|_, _|_, _|_, s(X), Y, Z, U, V, W, Q) > f(X, X, X, X, Y, Z, U, V, W, Q) because [23], by definition 23] f*(_|_, _|_, _|_, s(X), Y, Z, U, V, W, Q) >= f(X, X, X, X, Y, Z, U, V, W, Q) because f in Mul, [24], [24], [24], [24], [8], [9], [10], [11], [12] and [13], by (Stat) 24] s(X) > X because [25], by definition 25] s*(X) >= X because [7], by (Select) 26] f(_|_, _|_, _|_, _|_, s(X), Y, Z, U, V, W) > f(X, X, X, X, X, Y, Z, U, V, W) because [27], by definition 27] f*(_|_, _|_, _|_, _|_, s(X), Y, Z, U, V, W) >= f(X, X, X, X, X, Y, Z, U, V, W) because f in Mul, [28], [28], [28], [28], [28], [9], [10], [11], [12] and [13], by (Stat) popout

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