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TRS Stand 20472 pair #381710068
details
property
value
status
complete
benchmark
boolean_rings.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n010.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0912530422211 seconds
cpu usage
0.087380312
max memory
4120576.0
stage attributes
key
value
output-size
6199
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. F : [] --> o T : [] --> o and : [o * o] --> o equiv : [o * o] --> o impl : [o * o] --> o neg : [o] --> o or : [o * o] --> o xor : [o * o] --> o xor(X, F) => X xor(X, neg(X)) => F and(X, T) => X and(X, F) => F and(X, X) => X and(xor(X, Y), Z) => xor(and(X, Z), and(Y, Z)) xor(X, X) => F impl(X, Y) => xor(and(X, Y), xor(X, T)) or(X, Y) => xor(and(X, Y), xor(X, Y)) equiv(X, Y) => xor(X, xor(Y, T)) neg(X) => xor(X, T) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): xor(X, F) >? X xor(X, neg(X)) >? F and(X, T) >? X and(X, F) >? F and(X, X) >? X and(xor(X, Y), Z) >? xor(and(X, Z), and(Y, Z)) xor(X, X) >? F impl(X, Y) >? xor(and(X, Y), xor(X, T)) or(X, Y) >? xor(and(X, Y), xor(X, Y)) equiv(X, Y) >? xor(X, xor(Y, T)) neg(X) >? xor(X, T) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[T]] = _|_ We choose Lex = {} and Mul = {F, and, equiv, impl, neg, or, xor}, and the following precedence: equiv > impl > neg > or > and > xor > F Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: xor(X, F) > X xor(X, neg(X)) >= F and(X, _|_) > X and(X, F) >= F and(X, X) >= X and(xor(X, Y), Z) > xor(and(X, Z), and(Y, Z)) xor(X, X) > F impl(X, Y) > xor(and(X, Y), xor(X, _|_)) or(X, Y) > xor(and(X, Y), xor(X, Y)) equiv(X, Y) >= xor(X, xor(Y, _|_)) neg(X) >= xor(X, _|_) With these choices, we have: 1] xor(X, F) > X because [2], by definition 2] xor*(X, F) >= X because [3], by (Select) 3] X >= X by (Meta) 4] xor(X, neg(X)) >= F because [5], by (Star) 5] xor*(X, neg(X)) >= F because xor > F, by (Copy) 6] and(X, _|_) > X because [7], by definition 7] and*(X, _|_) >= X because [3], by (Select) 8] and(X, F) >= F because [9], by (Star) 9] and*(X, F) >= F because and > F, by (Copy) 10] and(X, X) >= X because [11], by (Star) 11] and*(X, X) >= X because [3], by (Select) 12] and(xor(X, Y), Z) > xor(and(X, Z), and(Y, Z)) because [13], by definition 13] and*(xor(X, Y), Z) >= xor(and(X, Z), and(Y, Z)) because and > xor, [14] and [18], by (Copy) 14] and*(xor(X, Y), Z) >= and(X, Z) because and in Mul, [15] and [17], by (Stat) 15] xor(X, Y) > X because [16], by definition 16] xor*(X, Y) >= X because [3], by (Select) 17] Z >= Z by (Meta) 18] and*(xor(X, Y), Z) >= and(Y, Z) because and in Mul, [19] and [17], by (Stat) 19] xor(X, Y) > Y because [20], by definition 20] xor*(X, Y) >= Y because [21], by (Select) 21] Y >= Y by (Meta)
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