C Integer pair #487569221

details

property	value
status	complete
benchmark	PastaA9.c
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	Stroeder_15

run statistics

property	value
solver	AProVE
configuration	c
runtime (wallclock)	9.63088393211 seconds
cpu usage	9.982377105
max memory	7.01956096E8

stage attributes

key	value
output-size	3489
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_c /export/starexec/sandbox2/benchmark/theBenchmark.c /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1413 ms]
(6) YES


----------------------------------------

(0)
Obligation:
c file /export/starexec/sandbox2/benchmark/theBenchmark.c
----------------------------------------

(1) CToIRSProof (EQUIVALENT)
Parsed C Integer Program as IRS.
----------------------------------------

(2)
Obligation:
Rules:
f1(x, y, z) -> f2(x_1, y, z) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE
f8(x9, x10, x11) -> f9(x9, x10, arith) :|: TRUE && arith = x11 + x10
f5(x12, x13, x14) -> f8(x12, x13, x14) :|: x12 >= x14
f9(x15, x16, x17) -> f5(x15, x16, x17) :|: TRUE
f5(x18, x19, x20) -> f10(x18, x19, x20) :|: x18 < x20
f4(x21, x22, x23) -> f5(x21, x22, x23) :|: x22 > 0
f4(x24, x25, x26) -> f6(x24, x25, x26) :|: x25 <= 0
f10(x27, x28, x29) -> f7(x27, x28, x29) :|: TRUE
f6(x30, x31, x32) -> f7(x30, x31, x32) :|: TRUE
Start term: f1(x, y, z)

----------------------------------------

(3) IRS2T2 (EQUIVALENT)
Transformed input IRS into an integer transition system.Used the following mapping from defined symbols to location IDs:

   (f1_3,1)
   (f2_3,2)
   (f3_3,3)
   (f4_3,4)
   (f8_3,5)
   (f9_3,6)
   (f5_3,7)
   (f10_3,8)
   (f6_3,9)
   (f7_3,10)

----------------------------------------

(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX3;
x1 := oldX1;
x2 := oldX2;
TO: 2;

FROM: 2;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX0;
x1 := oldX3;
x2 := oldX2;
TO: 3;

FROM: 3;
oldX0 := x0;
oldX1 := x1;
oldX2 := x2;
oldX3 := nondet();
assume(0 = 0);
x0 := oldX0;

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actions all output return to C Integer