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TRS Stand 20472 pair #381711864
details
property
value
status
complete
benchmark
ternary.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n080.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.293437004089 seconds
cpu usage
0.289452395
max memory
1.1833344E7
stage attributes
key
value
output-size
14316
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. !940 : [] --> o !minus : [o * o] --> o !plus : [o * o] --> o !times : [o * o] --> o 0 : [o] --> o 1 : [o] --> o j : [o] --> o opp : [o] --> o 0(!940) => !940 !plus(!940, X) => X !plus(X, !940) => X !plus(0(X), 0(Y)) => 0(!plus(X, Y)) !plus(0(X), 1(Y)) => 1(!plus(X, Y)) !plus(1(X), 0(Y)) => 1(!plus(X, Y)) !plus(0(X), j(Y)) => j(!plus(X, Y)) !plus(j(X), 0(Y)) => j(!plus(X, Y)) !plus(1(X), 1(Y)) => j(!plus(!plus(X, Y), 1(!940))) !plus(j(X), j(Y)) => 1(!plus(!plus(X, Y), j(!940))) !plus(1(X), j(Y)) => 0(!plus(X, Y)) !plus(j(X), 1(Y)) => 0(!plus(X, Y)) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) opp(!940) => !940 opp(0(X)) => 0(opp(X)) opp(1(X)) => j(opp(X)) opp(j(X)) => 1(opp(X)) !minus(X, Y) => !plus(X, opp(Y)) !times(!940, X) => !940 !times(0(X), Y) => 0(!times(X, Y)) !times(1(X), Y) => !plus(0(!times(X, Y)), Y) !times(j(X), Y) => !minus(0(!times(X, Y)), Y) !times(!times(X, Y), Z) => !times(X, !times(Y, Z)) 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] !plus#(0(X), 0(Y)) =#> 0#(!plus(X, Y)) 1] !plus#(0(X), 0(Y)) =#> !plus#(X, Y) 2] !plus#(0(X), 1(Y)) =#> !plus#(X, Y) 3] !plus#(1(X), 0(Y)) =#> !plus#(X, Y) 4] !plus#(0(X), j(Y)) =#> !plus#(X, Y) 5] !plus#(j(X), 0(Y)) =#> !plus#(X, Y) 6] !plus#(1(X), 1(Y)) =#> !plus#(!plus(X, Y), 1(!940)) 7] !plus#(1(X), 1(Y)) =#> !plus#(X, Y) 8] !plus#(j(X), j(Y)) =#> !plus#(!plus(X, Y), j(!940)) 9] !plus#(j(X), j(Y)) =#> !plus#(X, Y) 10] !plus#(1(X), j(Y)) =#> 0#(!plus(X, Y)) 11] !plus#(1(X), j(Y)) =#> !plus#(X, Y) 12] !plus#(j(X), 1(Y)) =#> 0#(!plus(X, Y)) 13] !plus#(j(X), 1(Y)) =#> !plus#(X, Y) 14] !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z)) 15] !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z) 16] opp#(0(X)) =#> 0#(opp(X)) 17] opp#(0(X)) =#> opp#(X) 18] opp#(1(X)) =#> opp#(X) 19] opp#(j(X)) =#> opp#(X) 20] !minus#(X, Y) =#> !plus#(X, opp(Y)) 21] !minus#(X, Y) =#> opp#(Y) 22] !times#(0(X), Y) =#> 0#(!times(X, Y)) 23] !times#(0(X), Y) =#> !times#(X, Y) 24] !times#(1(X), Y) =#> !plus#(0(!times(X, Y)), Y) 25] !times#(1(X), Y) =#> 0#(!times(X, Y)) 26] !times#(1(X), Y) =#> !times#(X, Y) 27] !times#(j(X), Y) =#> !minus#(0(!times(X, Y)), Y) 28] !times#(j(X), Y) =#> 0#(!times(X, Y)) 29] !times#(j(X), Y) =#> !times#(X, Y) 30] !times#(!times(X, Y), Z) =#> !times#(X, !times(Y, Z)) 31] !times#(!times(X, Y), Z) =#> !times#(Y, Z) Rules R_0: 0(!940) => !940 !plus(!940, X) => X !plus(X, !940) => X !plus(0(X), 0(Y)) => 0(!plus(X, Y)) !plus(0(X), 1(Y)) => 1(!plus(X, Y)) !plus(1(X), 0(Y)) => 1(!plus(X, Y)) !plus(0(X), j(Y)) => j(!plus(X, Y)) !plus(j(X), 0(Y)) => j(!plus(X, Y)) !plus(1(X), 1(Y)) => j(!plus(!plus(X, Y), 1(!940))) !plus(j(X), j(Y)) => 1(!plus(!plus(X, Y), j(!940))) !plus(1(X), j(Y)) => 0(!plus(X, Y)) !plus(j(X), 1(Y)) => 0(!plus(X, Y)) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) opp(!940) => !940 opp(0(X)) => 0(opp(X))
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