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C Integ Progr 58257 pair #381736870
details
property
value
status
complete
benchmark
WhileDecr.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n096.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
4.03733491898 seconds
cpu usage
7.944251114
max memory
3.89881856E8
stage attributes
key
value
output-size
2269
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.c # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) IRS2T2 [EQUIVALENT, 0 ms] (4) T2IntSys (5) T2 [EQUIVALENT, 1043 ms] (6) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(i) -> f2(x_1) :|: TRUE f3(x) -> f4(arith) :|: TRUE && arith = x - 1 f2(x1) -> f3(x1) :|: x1 > 5 f4(x2) -> f2(x2) :|: TRUE f2(x3) -> f5(x3) :|: x3 <= 5 Start term: f1(i) ---------------------------------------- (3) IRS2T2 (EQUIVALENT) Transformed input IRS into an integer transition system.Used the following mapping from defined symbols to location IDs: (f1_1,1) (f2_1,2) (f3_1,3) (f4_1,4) (f5_1,5) ---------------------------------------- (4) Obligation: START: 1; FROM: 1; oldX0 := x0; oldX1 := nondet(); assume(0 = 0); x0 := oldX1; TO: 2; FROM: 3; oldX0 := x0; oldX1 := -(1 - oldX0); assume(0 = 0 && oldX1 = oldX0 - 1); x0 := -(1 - oldX0); TO: 4; FROM: 2; oldX0 := x0; assume(oldX0 > 5); x0 := oldX0; TO: 3; FROM: 4; oldX0 := x0; assume(0 = 0); x0 := oldX0; TO: 2; FROM: 2; oldX0 := x0; assume(oldX0 <= 5); x0 := oldX0; TO: 5; ---------------------------------------- (5) T2 (EQUIVALENT) Initially, performed program simplifications using lexicographic rank functions: * Removed transitions 4, 7, 8 using the following rank functions: - Rank function 1: RF for loc. 6: 1+2*x0
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