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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 42 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) TerminationGraphProcessor [EQUIVALENT, 3 ms]
(8) IntTRS
(9) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(10) IntTRS
(11) RankingReductionPairProof [EQUIVALENT, 0 ms]
(12) 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(x, y) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 - 1
f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = 0 - 2 * x15
f3(x8, x9) -> f4(x8, x9) :|: x8 + x9 > 0
f6(x10, x11) -> f3(x10, x11) :|: TRUE
f3(x12, x13) -> f7(x12, x13) :|: x12 + x13 <= 0
Start term: f1(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:
f3(x8, x9) -> f4(x8, x9) :|: x8 + x9 > 0
f6(x10, x11) -> f3(x10, x11) :|: TRUE
f5(x14, x15) -> f6(x14, x16) :|: TRUE && x16 = 0 - 2 * x15
f4(x4, x5) -> f5(arith, x5) :|: TRUE && arith = x4 - 1

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x14:0, x15:0) -> f5(x14:0 - 1, 0 - 2 * x15:0) :|: x14:0 + (0 - 2 * x15:0) > 0

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(7) TerminationGraphProcessor (EQUIVALENT)
Constructed the termination graph and obtained one non-trivial SCC.

f5(x14:0, x15:0) -> f5(x14:0 - 1, 0 - 2 * x15:0) :|: x14:0 + (0 - 2 * x15:0) > 0
has been transformed into
f5(x14:0, x15:0) -> f5(x14:0 - 1, 0 - 2 * x15:0) :|: x15:0 = 0 - 2 * x5 && (x14:0 = x4 - 1 && x14:0 + (0 - 2 * x15:0) > 0) && x4 + (0 - 2 * x5) > 0.


f5(x14:0, x15:0) -> f5(x14:0 - 1, 0 - 2 * x15:0) :|: x15:0 = 0 - 2 * x5 && (x14:0 = x4 - 1 && x14:0 + (0 - 2 * x15:0) > 0) && x4 + (0 - 2 * x5) > 0 and 
f5(x14:0, x15:0) -> f5(x14:0 - 1, 0 - 2 * x15:0) :|: x15:0 = 0 - 2 * x5 && (x14:0 = x4 - 1 && x14:0 + (0 - 2 * x15:0) > 0) && x4 + (0 - 2 * x5) > 0
have been merged into the new rule
f5(x14, x15) -> f5(x14 - 1 - 1, 0 - 2 * (0 - 2 * x15)) :|: x15 = 0 - 2 * x16 && (x14 = x17 - 1 && x14 + (0 - 2 * x15) > 0) && x17 + (0 - 2 * x16) > 0 && (0 - 2 * x15 = 0 - 2 * x18 && (x14 - 1 = x19 - 1 && x14 - 1 + (0 - 2 * (0 - 2 * x15)) > 0) && x19 + (0 - 2 * x18) > 0)


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(8)
Obligation:
Rules:
f5(x20, x21) -> f5(x20 + -2, 4 * x21) :|: TRUE && x21 + 2 * x22 = 0 && x20 + -1 * x23 = -1 && x20 + -2 * x21 >= 1 && x23 + -2 * x22 >= 1 && -1 * x21 + x24 = 0 && x20 + -1 * x25 = 0 && x20 + 4 * x21 >= 2 && x25 + -2 * x24 >= 1

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(9) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(10)
Obligation:
Rules:
f5(x20:0, x21:0) -> f5(x20:0 - 2, 4 * x21:0) :|: x20:0 + 4 * x21:0 >= 2 && x25:0 + -2 * x24:0 >= 1 && x20:0 + -1 * x25:0 = 0 && 0 = -1 * x21:0 + x24:0 && x23:0 + -2 * x22:0 >= 1 && x20:0 + -2 * x21:0 >= 1 && x21:0 + 2 * x22:0 = 0 && x20:0 + -1 * x23:0 = -1

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(11) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f5 ] = 1/2*f5_1

The following rules are decreasing:
f5(x20:0, x21:0) -> f5(x20:0 - 2, 4 * x21:0) :|: x20:0 + 4 * x21:0 >= 2 && x25:0 + -2 * x24:0 >= 1 && x20:0 + -1 * x25:0 = 0 && 0 = -1 * x21:0 + x24:0 && x23:0 + -2 * x22:0 >= 1 && x20:0 + -2 * x21:0 >= 1 && x21:0 + 2 * x22:0 = 0 && x20:0 + -1 * x23:0 = -1

The following rules are bounded:
f5(x20:0, x21:0) -> f5(x20:0 - 2, 4 * x21:0) :|: x20:0 + 4 * x21:0 >= 2 && x25:0 + -2 * x24:0 >= 1 && x20:0 + -1 * x25:0 = 0 && 0 = -1 * x21:0 + x24:0 && x23:0 + -2 * x22:0 >= 1 && x20:0 + -2 * x21:0 >= 1 && x21:0 + 2 * x22:0 = 0 && x20:0 + -1 * x23:0 = -1


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(12)
YES