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C_Integer pair #487095567
details
property
value
status
complete
benchmark
Nyala-2lex_true-termination.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n149.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
4.23555 seconds
cpu usage
8.33186
user time
8.02702
system time
0.304834
max virtual memory
1.8815216E7
max residence set size
547552.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.c # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) IRS2T2 [EQUIVALENT, 0 ms] (4)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, 1113 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: f2(x, y) -> f3(x, arith) :|: TRUE && arith = y - 1 f4(x20, x21) -> f7(x22, x21) :|: TRUE && x22 = x20 - 1 f7(x3, x4) -> f8(x3, x5) :|: TRUE f3(x6, x7) -> f4(x6, x7) :|: x7 < 0 f3(x8, x9) -> f5(x8, x9) :|: x9 >= 0 f8(x10, x11) -> f6(x10, x11) :|: TRUE f5(x12, x13) -> f6(x12, x13) :|: TRUE f1(x14, x15) -> f2(x14, x15) :|: x14 >= 0 && x15 >= 0 f6(x16, x17) -> f1(x16, x17) :|: TRUE f1(x18, x19) -> f9(x18, x19) :|: x18 < 0 f1(x23, x24) -> f9(x23, x24) :|: x24 < 0 Start term: f1(x, y) ---------------------------------------- (3) IRS2T2 (EQUIVALENT) Transformed input IRS into an integer transition system.Used the following mapping from defined symbols to location IDs: (f2_2,1) (f3_2,2) (f4_2,3) (f7_2,4) (f8_2,5) (f5_2,6) (f6_2,7) (f1_2,8) (f9_2,9) ---------------------------------------- (4) Obligation: START: 8; FROM: 1; oldX0 := x0; oldX1 := x1; oldX2 := -(1 - oldX1); assume(0 = 0 && oldX2 = oldX1 - 1); x0 := oldX0; x1 := -(1 - oldX1); TO: 2; FROM: 3; oldX0 := x0; oldX1 := x1; oldX2 := -(1 - oldX0); assume(0 = 0 && oldX2 = oldX0 - 1); x0 := -(1 - oldX0); x1 := oldX1; TO: 4; FROM: 4; oldX0 := x0; oldX1 := x1; oldX2 := nondet(); assume(0 = 0); x0 := oldX0; x1 := oldX2; TO: 5; FROM: 2; oldX0 := x0; oldX1 := x1; assume(oldX1 < 0); x0 := oldX0; x1 := oldX1; TO: 3; FROM: 2;
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