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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 41 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 21 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 9 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(10) 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(q, y) -> f2(x_1, y) :|: TRUE
f2(x, x1) -> f3(x, x2) :|: TRUE
f5(x3, x4) -> f8(arith, x4) :|: TRUE && arith = x3 - x4 - 1
f6(x21, x22) -> f9(x23, x22) :|: TRUE && x23 = x21 + x22 - 1
f4(x7, x8) -> f5(x7, x8) :|: x8 > 0
f4(x9, x10) -> f6(x9, x10) :|: x10 <= 0
f8(x11, x12) -> f7(x11, x12) :|: TRUE
f9(x13, x14) -> f7(x13, x14) :|: TRUE
f3(x15, x16) -> f4(x15, x16) :|: x15 > 0
f7(x17, x18) -> f3(x17, x18) :|: TRUE
f3(x19, x20) -> f10(x19, x20) :|: x19 <= 0
Start term: f1(q, 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(x15, x16) -> f4(x15, x16) :|: x15 > 0
f7(x17, x18) -> f3(x17, x18) :|: TRUE
f8(x11, x12) -> f7(x11, x12) :|: TRUE
f5(x3, x4) -> f8(arith, x4) :|: TRUE && arith = x3 - x4 - 1
f4(x7, x8) -> f5(x7, x8) :|: x8 > 0
f9(x13, x14) -> f7(x13, x14) :|: TRUE
f6(x21, x22) -> f9(x23, x22) :|: TRUE && x23 = x21 + x22 - 1
f4(x9, x10) -> f6(x9, x10) :|: x10 <= 0

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1
f7(x, x1) -> f7(x - x1 - 1, x1) :|: x > 0 && x1 > 0

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f7(x, x1)] = -2 + x + x1

The following rules are decreasing:
f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1
f7(x, x1) -> f7(x - x1 - 1, x1) :|: x > 0 && x1 > 0
The following rules are bounded:
f7(x, x1) -> f7(x - x1 - 1, x1) :|: x > 0 && x1 > 0

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(8)
Obligation:
Rules:
f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f7(x, x1)] = x

The following rules are decreasing:
f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1
The following rules are bounded:
f7(x17:0, x18:0) -> f7(x17:0 + x18:0 - 1, x18:0) :|: x17:0 > 0 && x18:0 < 1

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(10)
YES