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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) TerminationGraphProcessor [SOUND, 45 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(8) IntTRS
(9) RankingReductionPairProof [EQUIVALENT, 25 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(c, x, y) -> f2(c, x_1, y) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x2, x4) :|: TRUE
f3(x5, x6, x7) -> f4(0, x6, x7) :|: TRUE
f5(x8, x9, x10) -> f6(x8, x9, arith) :|: TRUE && arith = x10 + 1
f6(x23, x24, x25) -> f7(x26, x24, x25) :|: TRUE && x26 = x23 + 1
f4(x14, x15, x16) -> f5(x14, x15, x16) :|: x15 > x16
f7(x17, x18, x19) -> f4(x17, x18, x19) :|: TRUE
f4(x20, x21, x22) -> f8(x20, x21, x22) :|: x21 <= x22
Start term: f1(c, x, y)

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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(x14, x15, x16) -> f5(x14, x15, x16) :|: x15 > x16
f7(x17, x18, x19) -> f4(x17, x18, x19) :|: TRUE
f6(x23, x24, x25) -> f7(x26, x24, x25) :|: TRUE && x26 = x23 + 1
f5(x8, x9, x10) -> f6(x8, x9, arith) :|: TRUE && arith = x10 + 1

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x23:0, x24:0, x25:0) -> f6(x23:0 + 1, x24:0, x25:0 + 1) :|: x25:0 < x24:0

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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(x2, x3)

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(8)
Obligation:
Rules:
f6(x24:0, x25:0) -> f6(x24:0, x25:0 + 1) :|: x25:0 < x24:0

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(9) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f6 ] = -1*f6_2 + f6_1

The following rules are decreasing:
f6(x24:0, x25:0) -> f6(x24:0, x25:0 + 1) :|: x25:0 < x24:0

The following rules are bounded:
f6(x24:0, x25:0) -> f6(x24:0, x25:0 + 1) :|: x25:0 < x24:0


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(10)
YES