/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.c
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given C Problem could be proven:

(0) C Problem
(1) CToIRSProof [EQUIVALENT, 0 ms]
(2) IntTRS
(3) TerminationGraphProcessor [SOUND, 50 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IntTRS
(7) TerminationGraphProcessor [EQUIVALENT, 2 ms]
(8) IntTRS
(9) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(10) IntTRS
(11) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(12) 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(q, z) -> f2(x_1, z) :|: TRUE
f2(x, x1) -> f3(x, x2) :|: TRUE
f4(x3, x4) -> f5(arith, x4) :|: TRUE && arith = x3 + x4 - 1
f5(x13, x14) -> f6(x13, x15) :|: TRUE && x15 = 0 - x14
f3(x7, x8) -> f4(x7, x8) :|: x7 > 0
f6(x9, x10) -> f3(x9, x10) :|: TRUE
f3(x11, x12) -> f7(x11, x12) :|: x11 <= 0
Start term: f1(q, z)

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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(x7, x8) -> f4(x7, x8) :|: x7 > 0
f6(x9, x10) -> f3(x9, x10) :|: TRUE
f5(x13, x14) -> f6(x13, x15) :|: TRUE && x15 = 0 - x14
f4(x3, x4) -> f5(arith, x4) :|: TRUE && arith = x3 + x4 - 1

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x13:0, x14:0) -> f5(x13:0 + (0 - x14:0) - 1, 0 - x14:0) :|: x13:0 > 0

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

f5(x13:0, x14:0) -> f5(x13:0 + (0 - x14:0) - 1, 0 - x14:0) :|: x13:0 > 0
has been transformed into
f5(x13:0, x14:0) -> f5(x13:0 + (0 - x14:0) - 1, 0 - x14:0) :|: x13:0 > 0 && x4 > 0.


f5(x13:0, x14:0) -> f5(x13:0 + (0 - x14:0) - 1, 0 - x14:0) :|: x13:0 > 0 && x4 > 0 and 
f5(x13:0, x14:0) -> f5(x13:0 + (0 - x14:0) - 1, 0 - x14:0) :|: x13:0 > 0 && x4 > 0
have been merged into the new rule
f5(x12, x13) -> f5(x12 + (0 - x13) - 1 + (0 - (0 - x13)) - 1, 0 - (0 - x13)) :|: x12 > 0 && x14 > 0 && (x12 + (0 - x13) - 1 > 0 && x15 > 0)


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(8)
Obligation:
Rules:
f5(x16, x17) -> f5(x16 + -2, x17) :|: TRUE && x16 >= 1 && x18 >= 1 && x16 + -1 * x17 >= 2 && x19 >= 1

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(9) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(10)
Obligation:
Rules:
f5(x16:0, x17:0) -> f5(x16:0 - 2, x17:0) :|: x16:0 + -1 * x17:0 >= 2 && x19:0 > 0 && x16:0 > 0 && x18:0 > 0

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(11) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x, x1)] = -1 + x - x1

The following rules are decreasing:
f5(x16:0, x17:0) -> f5(x16:0 - 2, x17:0) :|: x16:0 + -1 * x17:0 >= 2 && x19:0 > 0 && x16:0 > 0 && x18:0 > 0
The following rules are bounded:
f5(x16:0, x17:0) -> f5(x16:0 - 2, x17:0) :|: x16:0 + -1 * x17:0 >= 2 && x19:0 > 0 && x16:0 > 0 && x18:0 > 0

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(12)
YES