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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) TerminationGraphProcessor [SOUND, 41 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) RankingReductionPairProof [EQUIVALENT, 16 ms]
(10) YES


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(0)
Obligation:
c file /export/starexec/sandbox2/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, c) -> f2(0, c) :|: TRUE
f2(x, x1) -> f3(x, 0) :|: TRUE
f4(x2, x3) -> f5(arith, x3) :|: TRUE && arith = x2 + 1
f5(x12, x13) -> f6(x12, x14) :|: TRUE && x14 = x13 + 1
f3(x6, x7) -> f4(x6, x7) :|: x6 <= 10
f6(x8, x9) -> f3(x8, x9) :|: TRUE
f3(x10, x11) -> f7(x10, x11) :|: x10 > 10
Start term: f1(i, c)

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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 <= 10
f6(x8, x9) -> f3(x8, x9) :|: TRUE
f5(x12, x13) -> f6(x12, x14) :|: TRUE && x14 = x13 + 1
f4(x2, x3) -> f5(arith, x3) :|: TRUE && arith = x2 + 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 < 11

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

   f5(x1, x2) -> f5(x1)

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(8)
Obligation:
Rules:
f5(x12:0) -> f5(x12:0 + 1) :|: x12:0 < 11

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(9) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f5 ] = -1*f5_1

The following rules are decreasing:
f5(x12:0) -> f5(x12:0 + 1) :|: x12:0 < 11

The following rules are bounded:
f5(x12:0) -> f5(x12:0 + 1) :|: x12:0 < 11


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(10)
YES