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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 29 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 8 ms]
(6) IntTRS
(7) TerminationGraphProcessor [EQUIVALENT, 19 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/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(a, b) -> f2(arith, b) :|: TRUE && arith = 0 - x_1
f2(x13, x14) -> f3(x13, x15) :|: TRUE && x15 = 0 - x16
f4(x17, x18) -> f5(x17, x19) :|: TRUE && x19 = x18 + x17
f5(x20, x21) -> f6(x22, x21) :|: TRUE && x22 = x20 + 1
f3(x7, x8) -> f4(x7, x8) :|: x7 > x8
f6(x9, x10) -> f3(x9, x10) :|: TRUE
f3(x11, x12) -> f7(x11, x12) :|: x11 <= x12
Start term: f1(a, b)

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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 > x8
f6(x9, x10) -> f3(x9, x10) :|: TRUE
f5(x20, x21) -> f6(x22, x21) :|: TRUE && x22 = x20 + 1
f4(x17, x18) -> f5(x17, x19) :|: TRUE && x19 = x18 + x17

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x20:0, x21:0) -> f5(x20:0 + 1, x21:0 + (x20:0 + 1)) :|: x21:0 < x20:0 + 1

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

f5(x20:0, x21:0) -> f5(x20:0 + 1, x21:0 + (x20:0 + 1)) :|: x21:0 < x20:0 + 1
has been transformed into
f5(x20:0, x21:0) -> f5(x20:0 + 1, x21:0 + (x20:0 + 1)) :|: x21:0 = x5 + (x4 + 1) && (x20:0 = x4 + 1 && x21:0 < x20:0 + 1) && x5 < x4 + 1.


f5(x20:0, x21:0) -> f5(x20:0 + 1, x21:0 + (x20:0 + 1)) :|: x21:0 = x5 + (x4 + 1) && (x20:0 = x4 + 1 && x21:0 < x20:0 + 1) && x5 < x4 + 1 and 
f5(x20:0, x21:0) -> f5(x20:0 + 1, x21:0 + (x20:0 + 1)) :|: x21:0 = x5 + (x4 + 1) && (x20:0 = x4 + 1 && x21:0 < x20:0 + 1) && x5 < x4 + 1
have been merged into the new rule
f5(x14, x15) -> f5(x14 + 1 + 1, x15 + (x14 + 1) + (x14 + 1 + 1)) :|: x15 = x16 + (x17 + 1) && (x14 = x17 + 1 && x15 < x14 + 1) && x16 < x17 + 1 && (x15 + (x14 + 1) = x18 + (x19 + 1) && (x14 + 1 = x19 + 1 && x15 + (x14 + 1) < x14 + 1 + 1) && x18 < x19 + 1)


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(8)
Obligation:
Rules:
f5(x20, x21) -> f5(x20 + 2, x21 + 2 * x20 + 3) :|: TRUE && x21 + -1 * x22 + -1 * x23 = 1 && x20 + -1 * x23 = 1 && x21 + -1 * x20 <= 0 && x22 + -1 * x23 <= 0 && x21 + x20 + -1 * x24 + -1 * x25 = 0 && x20 + -1 * x25 = 0 && x21 <= 0 && x24 + -1 * x25 <= 0

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(9) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(10)
Obligation:
Rules:
f5(x20:0, x21:0) -> f5(x20:0 + 2, x21:0 + 2 * x20:0 + 3) :|: x21:0 < 1 && x24:0 + -1 * x25:0 <= 0 && x20:0 + -1 * x25:0 = 0 && x21:0 + x20:0 + -1 * x24:0 + -1 * x25:0 = 0 && x22:0 + -1 * x23:0 <= 0 && x21:0 + -1 * x20:0 <= 0 && x21:0 + -1 * x22:0 + -1 * x23:0 = 1 && x20:0 + -1 * x23:0 = 1

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(11) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f5(x, x1)] = x^2 - 2*x1

The following rules are decreasing:
f5(x20:0, x21:0) -> f5(x20:0 + 2, x21:0 + 2 * x20:0 + 3) :|: x21:0 < 1 && x24:0 + -1 * x25:0 <= 0 && x20:0 + -1 * x25:0 = 0 && x21:0 + x20:0 + -1 * x24:0 + -1 * x25:0 = 0 && x22:0 + -1 * x23:0 <= 0 && x21:0 + -1 * x20:0 <= 0 && x21:0 + -1 * x22:0 + -1 * x23:0 = 1 && x20:0 + -1 * x23:0 = 1
The following rules are bounded:
f5(x20:0, x21:0) -> f5(x20:0 + 2, x21:0 + 2 * x20:0 + 3) :|: x21:0 < 1 && x24:0 + -1 * x25:0 <= 0 && x20:0 + -1 * x25:0 = 0 && x21:0 + x20:0 + -1 * x24:0 + -1 * x25:0 = 0 && x22:0 + -1 * x23:0 <= 0 && x21:0 + -1 * x20:0 <= 0 && x21:0 + -1 * x22:0 + -1 * x23:0 = 1 && x20:0 + -1 * x23:0 = 1

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(12)
YES