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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 50 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 32 ms]
(6) IntTRS
(7) TerminationGraphProcessor [EQUIVALENT, 0 ms]
(8) IntTRS
(9) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(10) IntTRS
(11) PolynomialOrderProcessor [EQUIVALENT, 3 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(x, y) -> f2(x_1, y) :|: TRUE
f2(x1, x2) -> f3(x1, x3) :|: TRUE
f4(x4, x5) -> f5(x5, x5) :|: TRUE
f5(x6, x7) -> f6(x6, arith) :|: TRUE && arith = x7 - 1
f3(x8, x9) -> f4(x8, x9) :|: x8 > 0
f6(x10, x11) -> f3(x10, x11) :|: TRUE
f3(x12, x13) -> f7(x12, x13) :|: x12 <= 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 > 0
f6(x10, x11) -> f3(x10, x11) :|: TRUE
f5(x6, x7) -> f6(x6, arith) :|: TRUE && arith = x7 - 1
f4(x4, x5) -> f5(x5, x5) :|: TRUE

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x6:0, x7:0) -> f5(x7:0 - 1, x7:0 - 1) :|: x6:0 > 0

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

f5(x6:0, x7:0) -> f5(x7:0 - 1, x7:0 - 1) :|: x6:0 > 0
has been transformed into
f5(x6:0, x7:0) -> f5(x7:0 - 1, x7:0 - 1) :|: x6:0 > 0.


f5(x6:0, x7:0) -> f5(x7:0 - 1, x7:0 - 1) :|: x6:0 > 0 and 
f5(x6:0, x7:0) -> f5(x7:0 - 1, x7:0 - 1) :|: x6:0 > 0
have been merged into the new rule
f5(x10, x11) -> f5(x11 - 1 - 1, x11 - 1 - 1) :|: x10 > 0 && x11 - 1 > 0


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(8)
Obligation:
Rules:
f5(x12, x13) -> f5(x13 + -2, x13 + -2) :|: TRUE && x12 >= 1 && x13 >= 2

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(9) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(10)
Obligation:
Rules:
f5(x12:0, x13:0) -> f5(x13:0 - 2, x13:0 - 2) :|: x13:0 > 1 && x12:0 > 0

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

The following rules are decreasing:
f5(x12:0, x13:0) -> f5(x13:0 - 2, x13:0 - 2) :|: x13:0 > 1 && x12:0 > 0
The following rules are bounded:
f5(x12:0, x13:0) -> f5(x13:0 - 2, x13:0 - 2) :|: x13:0 > 1 && x12:0 > 0

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(12)
YES