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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 46 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 37 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 7 ms]
(8) IntTRS
(9) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms]
(10) IntTRS
(11) PolynomialOrderProcessor [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
f5(x4, x5) -> f6(x4, arith) :|: TRUE && arith = x5 - 1
f4(x6, x7) -> f5(x6, x7) :|: x7 > 0
f6(x8, x9) -> f4(x8, x9) :|: TRUE
f4(x10, x11) -> f7(x10, x11) :|: x11 <= 0
f7(x20, x21) -> f8(x22, x21) :|: TRUE && x22 = x20 - 1
f3(x14, x15) -> f4(x14, x15) :|: x14 > 0
f8(x16, x17) -> f3(x16, x17) :|: TRUE
f3(x18, x19) -> f9(x18, x19) :|: x18 <= 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(x14, x15) -> f4(x14, x15) :|: x14 > 0
f8(x16, x17) -> f3(x16, x17) :|: TRUE
f7(x20, x21) -> f8(x22, x21) :|: TRUE && x22 = x20 - 1
f4(x10, x11) -> f7(x10, x11) :|: x11 <= 0
f6(x8, x9) -> f4(x8, x9) :|: TRUE
f5(x4, x5) -> f6(x4, arith) :|: TRUE && arith = x5 - 1
f4(x6, x7) -> f5(x6, x7) :|: x7 > 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f4(x6:0, x7:0) -> f4(x6:0, x7:0 - 1) :|: x7:0 > 0
f4(x10:0, x11:0) -> f4(x10:0 - 1, x11:0) :|: x11:0 < 1 && x10:0 > 1

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f4(x, x1)] = -1 + x

The following rules are decreasing:
f4(x10:0, x11:0) -> f4(x10:0 - 1, x11:0) :|: x11:0 < 1 && x10:0 > 1
The following rules are bounded:
f4(x10:0, x11:0) -> f4(x10:0 - 1, x11:0) :|: x11:0 < 1 && x10:0 > 1

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(8)
Obligation:
Rules:
f4(x6:0, x7:0) -> f4(x6:0, x7:0 - 1) :|: x7:0 > 0

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

   f4(x1, x2) -> f4(x2)

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(10)
Obligation:
Rules:
f4(x7:0) -> f4(x7:0 - 1) :|: x7:0 > 0

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(11) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f4(x)] = x

The following rules are decreasing:
f4(x7:0) -> f4(x7:0 - 1) :|: x7:0 > 0
The following rules are bounded:
f4(x7:0) -> f4(x7:0 - 1) :|: x7:0 > 0

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(12)
YES