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TRS Stand 20472 pair #381712385
details
property
value
status
complete
benchmark
t012.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n093.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.150810003281 seconds
cpu usage
0.13503376
max memory
6279168.0
stage attributes
key
value
output-size
7923
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. !plus : [o * o] --> o !times : [o * o] --> o f : [o] --> o minus : [o] --> o minus(minus(X)) => X minus(!plus(X, Y)) => !times(minus(minus(minus(X))), minus(minus(minus(Y)))) minus(!times(X, Y)) => !plus(minus(minus(minus(X))), minus(minus(minus(Y)))) f(minus(X)) => minus(minus(minus(f(X)))) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] minus#(!plus(X, Y)) =#> minus#(minus(minus(X))) 1] minus#(!plus(X, Y)) =#> minus#(minus(X)) 2] minus#(!plus(X, Y)) =#> minus#(X) 3] minus#(!plus(X, Y)) =#> minus#(minus(minus(Y))) 4] minus#(!plus(X, Y)) =#> minus#(minus(Y)) 5] minus#(!plus(X, Y)) =#> minus#(Y) 6] minus#(!times(X, Y)) =#> minus#(minus(minus(X))) 7] minus#(!times(X, Y)) =#> minus#(minus(X)) 8] minus#(!times(X, Y)) =#> minus#(X) 9] minus#(!times(X, Y)) =#> minus#(minus(minus(Y))) 10] minus#(!times(X, Y)) =#> minus#(minus(Y)) 11] minus#(!times(X, Y)) =#> minus#(Y) 12] f#(minus(X)) =#> minus#(minus(minus(f(X)))) 13] f#(minus(X)) =#> minus#(minus(f(X))) 14] f#(minus(X)) =#> minus#(f(X)) 15] f#(minus(X)) =#> f#(X) Rules R_0: minus(minus(X)) => X minus(!plus(X, Y)) => !times(minus(minus(minus(X))), minus(minus(minus(Y)))) minus(!times(X, Y)) => !plus(minus(minus(minus(X))), minus(minus(minus(Y)))) f(minus(X)) => minus(minus(minus(f(X)))) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 4 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 9 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 11 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 12 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 13 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 14 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 15 : 12, 13, 14, 15 This graph has the following strongly connected components: P_1: minus#(!plus(X, Y)) =#> minus#(minus(minus(X))) minus#(!plus(X, Y)) =#> minus#(minus(X)) minus#(!plus(X, Y)) =#> minus#(X) minus#(!plus(X, Y)) =#> minus#(minus(minus(Y))) minus#(!plus(X, Y)) =#> minus#(minus(Y)) minus#(!plus(X, Y)) =#> minus#(Y) minus#(!times(X, Y)) =#> minus#(minus(minus(X))) minus#(!times(X, Y)) =#> minus#(minus(X)) minus#(!times(X, Y)) =#> minus#(X) minus#(!times(X, Y)) =#> minus#(minus(minus(Y))) minus#(!times(X, Y)) =#> minus#(minus(Y)) minus#(!times(X, Y)) =#> minus#(Y) P_2: f#(minus(X)) =#> f#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative).
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