details

property	value
status	complete
benchmark	#3.17a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n016.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.30641913414 seconds
cpu usage	0.303481382
max memory	9256960.0

stage attributes

key	value
output-size	9732
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. 0 : [] --> o app : [o * o] --> o cons : [o * o] --> o nil : [] --> o plus : [o * o] --> o pred : [o] --> o s : [o] --> o sum : [o] --> o app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) sum(plus(cons(0, X), cons(Y, Z))) => pred(sum(cons(s(X), cons(Y, Z)))) pred(cons(s(X), nil)) => cons(X, nil) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(nil, X) >? X app(X, nil) >? X app(cons(X, Y), Z) >? cons(X, app(Y, Z)) sum(cons(X, nil)) >? cons(X, nil) sum(cons(X, cons(Y, Z))) >? sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) >? sum(app(X, sum(cons(Y, cons(Z, U))))) plus(0, X) >? X plus(s(X), Y) >? s(plus(X, Y)) sum(plus(cons(0, X), cons(Y, Z))) >? pred(sum(cons(s(X), cons(Y, Z)))) pred(cons(s(X), nil)) >? cons(X, nil) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 app = \y0y1.y1 + 3y0 cons = \y0y1.y0 + y1 nil = 0 plus = \y0y1.y0 + y1 pred = \y0.y0 s = \y0.y0 sum = \y0.y0 Using this interpretation, the requirements translate to: [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] [[app(cons(_x0, _x1), _x2)]] = x2 + 3x0 + 3x1 >= x0 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] [[sum(cons(_x0, nil))]] = x0 >= x0 = [[cons(_x0, nil)]] [[sum(cons(_x0, cons(_x1, _x2)))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[sum(cons(plus(_x0, _x1), _x2))]] [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = x1 + x2 + x3 + 3x0 >= x1 + x2 + x3 + 3x0 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] [[plus(0, _x0)]] = 3 + x0 > x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[s(plus(_x0, _x1))]] [[sum(plus(cons(0, _x0), cons(_x1, _x2)))]] = 3 + x0 + x1 + x2 > x0 + x1 + x2 = [[pred(sum(cons(s(_x0), cons(_x1, _x2))))]] [[pred(cons(s(_x0), nil))]] = x0 >= x0 = [[cons(_x0, nil)]] We can thus remove the following rules: plus(0, X) => X sum(plus(cons(0, X), cons(Y, Z))) => pred(sum(cons(s(X), cons(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]): app(nil, X) >? X app(X, nil) >? X app(cons(X, Y), Z) >? cons(X, app(Y, Z)) sum(cons(X, nil)) >? cons(X, nil) sum(cons(X, cons(Y, Z))) >? sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) >? sum(app(X, sum(cons(Y, cons(Z, U))))) plus(s(X), Y) >? s(plus(X, Y)) pred(cons(s(X), nil)) >? cons(X, nil) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.y1 + 3y0 cons = \y0y1.y0 + y1 nil = 1 plus = \y0y1.y0 + y1 pred = \y0.3 + 3y0 s = \y0.y0 popout

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