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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 39 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 14 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 0 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(x, y, z) -> f2(x_1, y, z) :|: TRUE
f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE
f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE
f5(x9, x10, x11) -> f6(x9, x10, arith) :|: TRUE && arith = x11 + x10
f4(x12, x13, x14) -> f5(x12, x13, x14) :|: x13 > 0 && x12 >= x14
f6(x15, x16, x17) -> f4(x15, x16, x17) :|: TRUE
f4(x18, x19, x20) -> f7(x18, x19, x20) :|: x19 <= 0
f4(x21, x22, x23) -> f7(x21, x22, x23) :|: x21 < x23
Start term: f1(x, y, 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:
f4(x12, x13, x14) -> f5(x12, x13, x14) :|: x13 > 0 && x12 >= x14
f6(x15, x16, x17) -> f4(x15, x16, x17) :|: TRUE
f5(x9, x10, x11) -> f6(x9, x10, arith) :|: TRUE && arith = x11 + x10

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f6(x15:0, x16:0, x17:0) -> f6(x15:0, x16:0, x17:0 + x16:0) :|: x16:0 > 0 && x17:0 <= x15:0

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f6(x, x1, x2)] = x - x2

The following rules are decreasing:
f6(x15:0, x16:0, x17:0) -> f6(x15:0, x16:0, x17:0 + x16:0) :|: x16:0 > 0 && x17:0 <= x15:0
The following rules are bounded:
f6(x15:0, x16:0, x17:0) -> f6(x15:0, x16:0, x17:0 + x16:0) :|: x16:0 > 0 && x17:0 <= x15:0

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(8)
YES