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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 53 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 39 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 12 ms]
(8) 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) -> f2(x_1) :|: TRUE
f4(x) -> f7(arith) :|: TRUE && arith = x + 1
f5(x11) -> f8(x12) :|: TRUE && x12 = x11 + 2
f3(x2) -> f4(x2) :|: x3 < 0
f3(x13) -> f4(x13) :|: x14 > 0
f3(x4) -> f5(x4) :|: x5 = 0
f7(x6) -> f6(x6) :|: TRUE
f8(x7) -> f6(x7) :|: TRUE
f2(x8) -> f3(x8) :|: x8 < 255
f6(x9) -> f2(x9) :|: TRUE
f2(x10) -> f9(x10) :|: x10 >= 255
Start term: f1(i)

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

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(4)
Obligation:
Rules:
f2(x8) -> f3(x8) :|: x8 < 255
f6(x9) -> f2(x9) :|: TRUE
f7(x6) -> f6(x6) :|: TRUE
f4(x) -> f7(arith) :|: TRUE && arith = x + 1
f3(x2) -> f4(x2) :|: x3 < 0
f3(x13) -> f4(x13) :|: x14 > 0
f8(x7) -> f6(x7) :|: TRUE
f5(x11) -> f8(x12) :|: TRUE && x12 = x11 + 2
f3(x4) -> f5(x4) :|: x5 = 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x9:0) -> f6(x9:0 + 1) :|: x9:0 < 255 && x3:0 < 0
f6(x) -> f6(x + 1) :|: x < 255 && x1 > 0
f6(x2) -> f6(x2 + 2) :|: x2 < 255

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f6(x)] = 254 - x

The following rules are decreasing:
f6(x9:0) -> f6(x9:0 + 1) :|: x9:0 < 255 && x3:0 < 0
f6(x) -> f6(x + 1) :|: x < 255 && x1 > 0
f6(x2) -> f6(x2 + 2) :|: x2 < 255
The following rules are bounded:
f6(x9:0) -> f6(x9:0 + 1) :|: x9:0 < 255 && x3:0 < 0
f6(x) -> f6(x + 1) :|: x < 255 && x1 > 0
f6(x2) -> f6(x2 + 2) :|: x2 < 255

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(8)
YES