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TRS Stand 20472 pair #381714833
details
property
value
status
complete
benchmark
2.41.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n042.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.056244134903 seconds
cpu usage
0.034891203
max memory
1720320.0
stage attributes
key
value
output-size
2934
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 f : [o * o] --> o g : [o * o] --> o nil : [] --> o norm : [o] --> o rem : [o * o] --> o s : [o] --> o norm(nil) => 0 norm(g(X, Y)) => s(norm(X)) f(X, nil) => g(nil, X) f(X, g(Y, Z)) => g(f(X, Y), Z) rem(nil, X) => nil rem(g(X, Y), 0) => g(X, Y) rem(g(X, Y), s(Z)) => rem(X, 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]): norm(nil) >? 0 norm(g(X, Y)) >? s(norm(X)) f(X, nil) >? g(nil, X) f(X, g(Y, Z)) >? g(f(X, Y), Z) rem(nil, X) >? nil rem(g(X, Y), 0) >? g(X, Y) rem(g(X, Y), s(Z)) >? rem(X, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 f = \y0y1.2 + 2y0 + 3y1 g = \y0y1.2 + y0 + 2y1 nil = 0 norm = \y0.y0 rem = \y0y1.3 + y1 + 2y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[norm(nil)]] = 0 >= 0 = [[0]] [[norm(g(_x0, _x1))]] = 2 + x0 + 2x1 > 1 + x0 = [[s(norm(_x0))]] [[f(_x0, nil)]] = 2 + 2x0 >= 2 + 2x0 = [[g(nil, _x0)]] [[f(_x0, g(_x1, _x2))]] = 8 + 2x0 + 3x1 + 6x2 > 4 + 2x0 + 2x2 + 3x1 = [[g(f(_x0, _x1), _x2)]] [[rem(nil, _x0)]] = 3 + x0 > 0 = [[nil]] [[rem(g(_x0, _x1), 0)]] = 7 + 2x0 + 4x1 > 2 + x0 + 2x1 = [[g(_x0, _x1)]] [[rem(g(_x0, _x1), s(_x2))]] = 8 + x2 + 2x0 + 4x1 > 3 + x2 + 2x0 = [[rem(_x0, _x2)]] We can thus remove the following rules: norm(g(X, Y)) => s(norm(X)) f(X, g(Y, Z)) => g(f(X, Y), Z) rem(nil, X) => nil rem(g(X, Y), 0) => g(X, Y) rem(g(X, Y), s(Z)) => rem(X, 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]): norm(nil) >? 0 f(X, nil) >? g(nil, X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 f = \y0y1.3 + y0 + 3y1 g = \y0y1.y0 + y1 nil = 0 norm = \y0.3 + y0 Using this interpretation, the requirements translate to: [[norm(nil)]] = 3 > 0 = [[0]] [[f(_x0, nil)]] = 3 + x0 > x0 = [[g(nil, _x0)]] We can thus remove the following rules: norm(nil) => 0 f(X, nil) => g(nil, X) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++
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