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TRS Stand 20472 pair #381714703
details
property
value
status
complete
benchmark
4.22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n082.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0658719539642 seconds
cpu usage
0.062286892
max memory
3170304.0
stage attributes
key
value
output-size
3667
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. and : [o * o] --> o not : [o] --> o or : [o * o] --> o not(and(X, Y)) => or(not(X), not(Y)) not(or(X, Y)) => and(not(X), not(Y)) and(X, or(Y, Z)) => or(and(X, Y), and(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]): not(and(X, Y)) >? or(not(X), not(Y)) not(or(X, Y)) >? and(not(X), not(Y)) and(X, or(Y, Z)) >? or(and(X, Y), and(X, Z)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {and, not, or}, and the following precedence: not > and > or With these choices, we have: 1] not(and(X, Y)) >= or(not(X), not(Y)) because [2], by (Star) 2] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [3] and [7], by (Copy) 3] not*(and(X, Y)) >= not(X) because not in Mul and [4], by (Stat) 4] and(X, Y) > X because [5], by definition 5] and*(X, Y) >= X because [6], by (Select) 6] X >= X by (Meta) 7] not*(and(X, Y)) >= not(Y) because not in Mul and [8], by (Stat) 8] and(X, Y) > Y because [9], by definition 9] and*(X, Y) >= Y because [10], by (Select) 10] Y >= Y by (Meta) 11] not(or(X, Y)) > and(not(X), not(Y)) because [12], by definition 12] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [13] and [16], by (Copy) 13] not*(or(X, Y)) >= not(X) because not in Mul and [14], by (Stat) 14] or(X, Y) > X because [15], by definition 15] or*(X, Y) >= X because [6], by (Select) 16] not*(or(X, Y)) >= not(Y) because not in Mul and [17], by (Stat) 17] or(X, Y) > Y because [18], by definition 18] or*(X, Y) >= Y because [10], by (Select) 19] and(X, or(Y, Z)) > or(and(X, Y), and(X, Z)) because [20], by definition 20] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [21] and [25], by (Copy) 21] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [22] and [23], by (Stat) 22] X >= X by (Meta) 23] or(Y, Z) > Y because [24], by definition 24] or*(Y, Z) >= Y because [10], by (Select) 25] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [22] and [26], by (Stat) 26] or(Y, Z) > Z because [27], by definition 27] or*(Y, Z) >= Z because [28], by (Select) 28] Z >= Z by (Meta) We can thus remove the following rules: not(or(X, Y)) => and(not(X), not(Y)) and(X, or(Y, Z)) => or(and(X, Y), and(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]): not(and(X, Y)) >? or(not(X), not(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.3 + 3y0 + 3y1 not = \y0.3y0 or = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[not(and(_x0, _x1))]] = 9 + 9x0 + 9x1 > 3x0 + 3x1 = [[or(not(_x0), not(_x1))]] We can thus remove the following rules: not(and(X, Y)) => or(not(X), not(Y)) 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 +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
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