/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 51 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(10) 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, sum, bound) -> f2(i, sum, x_1) :|: TRUE
f2(x, x1, x2) -> f3(0, x1, x2) :|: TRUE
f3(x3, x4, x5) -> f4(x3, 0, x5) :|: TRUE
f5(x6, x7, x8) -> f6(x6, arith, x8) :|: TRUE && arith = x7 + x6
f6(x21, x22, x23) -> f7(x24, x22, x23) :|: TRUE && x24 = x21 + 1
f4(x12, x13, x14) -> f5(x12, x13, x14) :|: x12 < x14
f7(x15, x16, x17) -> f4(x15, x16, x17) :|: TRUE
f4(x18, x19, x20) -> f8(x18, x19, x20) :|: x18 >= x20
Start term: f1(i, sum, bound)

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

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(4)
Obligation:
Rules:
f4(x12, x13, x14) -> f5(x12, x13, x14) :|: x12 < x14
f7(x15, x16, x17) -> f4(x15, x16, x17) :|: TRUE
f6(x21, x22, x23) -> f7(x24, x22, x23) :|: TRUE && x24 = x21 + 1
f5(x6, x7, x8) -> f6(x6, arith, x8) :|: TRUE && arith = x7 + x6

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x21:0, x22:0, x23:0) -> f6(x21:0 + 1, x22:0 + (x21:0 + 1), x23:0) :|: x23:0 > x21:0 + 1

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(7) IntTRSUnneededArgumentFilterProof (EQUIVALENT)
Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements:

   f6(x1, x2, x3) -> f6(x1, x3)

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(8)
Obligation:
Rules:
f6(x21:0, x23:0) -> f6(x21:0 + 1, x23:0) :|: x23:0 > x21:0 + 1

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f6(x, x1)] = -x + x1

The following rules are decreasing:
f6(x21:0, x23:0) -> f6(x21:0 + 1, x23:0) :|: x23:0 > x21:0 + 1
The following rules are bounded:
f6(x21:0, x23:0) -> f6(x21:0 + 1, x23:0) :|: x23:0 > x21:0 + 1

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(10)
YES