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TRS Stand 20472 pair #381711879
details
property
value
status
complete
benchmark
test830.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n046.star.cs.uiowa.edu
space
Strategy_removed_mixed_05
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0761861801147 seconds
cpu usage
0.060755076
max memory
1990656.0
stage attributes
key
value
output-size
4939
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 cons : [o * o] --> o f : [o] --> o g : [o] --> o h : [o] --> o s : [o] --> o f(s(X)) => f(X) g(cons(0, X)) => g(X) g(cons(s(X), Y)) => s(X) h(cons(X, Y)) => h(g(cons(X, Y))) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> oa cons : [oa * oa] --> oa f : [oa] --> l g : [oa] --> oa h : [oa] --> pa s : [oa] --> oa 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)) >? f(X) g(cons(0, X)) >? g(X) g(cons(s(X), Y)) >? s(X) h(cons(X, Y)) >? h(g(cons(X, Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.y0 + y1 f = \y0.y0 g = \y0.y0 h = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[f(s(_x0))]] = x0 >= x0 = [[f(_x0)]] [[g(cons(0, _x0))]] = 3 + x0 > x0 = [[g(_x0)]] [[g(cons(s(_x0), _x1))]] = x0 + x1 >= x0 = [[s(_x0)]] [[h(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[h(g(cons(_x0, _x1)))]] We can thus remove the following rules: g(cons(0, X)) => g(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]): f(s(X)) >? f(X) g(cons(s(X), Y)) >? s(X) h(cons(X, Y)) >? h(g(cons(X, Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.1 + y0 + y1 f = \y0.y0 g = \y0.y0 h = \y0.2y0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[f(s(_x0))]] = 2x0 >= x0 = [[f(_x0)]] [[g(cons(s(_x0), _x1))]] = 1 + x1 + 2x0 > 2x0 = [[s(_x0)]] [[h(cons(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[h(g(cons(_x0, _x1)))]] We can thus remove the following rules: g(cons(s(X), Y)) => s(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]): f(s(X)) >? f(X) h(cons(X, Y)) >? h(g(cons(X, Y))) We orient these requirements with a polynomial interpretation in the natural numbers.
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