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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) TerminationGraphProcessor [SOUND, 41 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 2 ms]
(8) YES


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(0)
Obligation:
c file /export/starexec/sandbox/benchmark/theBenchmark.c
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(1) CToIRSProof (EQUIVALENT)
Parsed C Integer Program as IRS.
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(2)
Obligation:
Rules:
f1(i, j) -> f2(i, 1) :|: TRUE
f2(x, x1) -> f3(10000, x1) :|: TRUE
f4(x2, x3) -> f5(x2, arith) :|: TRUE && arith = x3 + 1
f5(x12, x13) -> f6(x14, x13) :|: TRUE && x14 = x12 - 1
f3(x6, x7) -> f4(x6, x7) :|: x6 - x7 >= 1
f6(x8, x9) -> f3(x8, x9) :|: TRUE
f3(x10, x11) -> f7(x10, x11) :|: x10 - x11 < 1
Start term: f1(i, j)

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(3) TerminationGraphProcessor (SOUND)
Constructed the termination graph and obtained one non-trivial SCC.

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(4)
Obligation:
Rules:
f3(x6, x7) -> f4(x6, x7) :|: x6 - x7 >= 1
f6(x8, x9) -> f3(x8, x9) :|: TRUE
f5(x12, x13) -> f6(x14, x13) :|: TRUE && x14 = x12 - 1
f4(x2, x3) -> f5(x2, arith) :|: TRUE && arith = x3 + 1

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x12:0, x13:0) -> f5(x12:0 - 1, x13:0 + 1) :|: x12:0 - 1 - x13:0 >= 1

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x, x1)] = -1 + x - x1

The following rules are decreasing:
f5(x12:0, x13:0) -> f5(x12:0 - 1, x13:0 + 1) :|: x12:0 - 1 - x13:0 >= 1
The following rules are bounded:
f5(x12:0, x13:0) -> f5(x12:0 - 1, x13:0 + 1) :|: x12:0 - 1 - x13:0 >= 1

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(8)
YES