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TRS Stand 20472 pair #381713313
details
property
value
status
complete
benchmark
gcd_triple.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n081.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.215620040894 seconds
cpu usage
0.205085073
max memory
8921088.0
stage attributes
key
value
output-size
8918
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. !minus : [o * o] --> o 0 : [] --> o gcd : [o * o * o] --> o max : [o * o] --> o min : [o * o] --> o s : [o] --> o min(X, 0) => 0 min(0, X) => 0 min(s(X), s(Y)) => s(min(X, Y)) max(X, 0) => X max(0, X) => X max(s(X), s(Y)) => s(max(X, Y)) !minus(X, 0) => X !minus(s(X), s(Y)) => !minus(X, Y) gcd(s(X), s(Y), Z) => gcd(!minus(max(X, Y), min(X, Y)), s(min(X, Y)), Z) gcd(X, s(Y), s(Z)) => gcd(X, !minus(max(Y, Z), min(Y, Z)), s(min(Y, Z))) gcd(s(X), Y, s(Z)) => gcd(!minus(max(X, Z), min(X, Z)), Y, s(min(X, Z))) gcd(X, 0, 0) => X gcd(0, X, 0) => X gcd(0, 0, X) => 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] min#(s(X), s(Y)) =#> min#(X, Y) 1] max#(s(X), s(Y)) =#> max#(X, Y) 2] !minus#(s(X), s(Y)) =#> !minus#(X, Y) 3] gcd#(s(X), s(Y), Z) =#> gcd#(!minus(max(X, Y), min(X, Y)), s(min(X, Y)), Z) 4] gcd#(s(X), s(Y), Z) =#> !minus#(max(X, Y), min(X, Y)) 5] gcd#(s(X), s(Y), Z) =#> max#(X, Y) 6] gcd#(s(X), s(Y), Z) =#> min#(X, Y) 7] gcd#(s(X), s(Y), Z) =#> min#(X, Y) 8] gcd#(X, s(Y), s(Z)) =#> gcd#(X, !minus(max(Y, Z), min(Y, Z)), s(min(Y, Z))) 9] gcd#(X, s(Y), s(Z)) =#> !minus#(max(Y, Z), min(Y, Z)) 10] gcd#(X, s(Y), s(Z)) =#> max#(Y, Z) 11] gcd#(X, s(Y), s(Z)) =#> min#(Y, Z) 12] gcd#(X, s(Y), s(Z)) =#> min#(Y, Z) 13] gcd#(s(X), Y, s(Z)) =#> gcd#(!minus(max(X, Z), min(X, Z)), Y, s(min(X, Z))) 14] gcd#(s(X), Y, s(Z)) =#> !minus#(max(X, Z), min(X, Z)) 15] gcd#(s(X), Y, s(Z)) =#> max#(X, Z) 16] gcd#(s(X), Y, s(Z)) =#> min#(X, Z) 17] gcd#(s(X), Y, s(Z)) =#> min#(X, Z) Rules R_0: min(X, 0) => 0 min(0, X) => 0 min(s(X), s(Y)) => s(min(X, Y)) max(X, 0) => X max(0, X) => X max(s(X), s(Y)) => s(max(X, Y)) !minus(X, 0) => X !minus(s(X), s(Y)) => !minus(X, Y) gcd(s(X), s(Y), Z) => gcd(!minus(max(X, Y), min(X, Y)), s(min(X, Y)), Z) gcd(X, s(Y), s(Z)) => gcd(X, !minus(max(Y, Z), min(Y, Z)), s(min(Y, Z))) gcd(s(X), Y, s(Z)) => gcd(!minus(max(X, Z), min(X, Z)), Y, s(min(X, Z))) gcd(X, 0, 0) => X gcd(0, X, 0) => X gcd(0, 0, X) => 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 : 1 * 2 : 2 * 3 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 4 : 2 * 5 : 1 * 6 : 0 * 7 : 0 * 8 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 9 : 2 * 10 : 1 * 11 : 0 * 12 : 0 * 13 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 14 : 2 * 15 : 1 * 16 : 0 * 17 : 0
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