/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(10) 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(x4, 1) :|: TRUE
f6(x6, x7) -> f7(x6, arith) :|: TRUE && arith = x7 + 1
f5(x8, x9) -> f6(x8, x9) :|: x9 < x8
f7(x10, x11) -> f5(x10, x11) :|: TRUE
f5(x12, x13) -> f8(x12, x13) :|: x13 >= x12
f8(x22, x23) -> f9(x24, x23) :|: TRUE && x24 = x22 - 1
f3(x16, x17) -> f4(x16, x17) :|: x16 >= 0
f9(x18, x19) -> f3(x18, x19) :|: TRUE
f3(x20, x21) -> f10(x20, x21) :|: x20 < 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(x16, x17) -> f4(x16, x17) :|: x16 >= 0
f9(x18, x19) -> f3(x18, x19) :|: TRUE
f8(x22, x23) -> f9(x24, x23) :|: TRUE && x24 = x22 - 1
f5(x12, x13) -> f8(x12, x13) :|: x13 >= x12
f4(x4, x5) -> f5(x4, 1) :|: TRUE
f7(x10, x11) -> f5(x10, x11) :|: TRUE
f6(x6, x7) -> f7(x6, arith) :|: TRUE && arith = x7 + 1
f5(x8, x9) -> f6(x8, x9) :|: x9 < x8

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x8:0, x9:0) -> f5(x8:0, x9:0 + 1) :|: x9:0 < x8:0
f5(x12:0, x13:0) -> f5(x12:0 - 1, 1) :|: x13:0 >= x12:0 && x12:0 > 0

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

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

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(8)
Obligation:
Rules:
f5(x8:0, x9:0) -> f5(x8:0, x9:0 + 1) :|: x9:0 < x8:0

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

The following rules are decreasing:
f5(x8:0, x9:0) -> f5(x8:0, x9:0 + 1) :|: x9:0 < x8:0
The following rules are bounded:
f5(x8:0, x9:0) -> f5(x8:0, x9:0 + 1) :|: x9:0 < x8:0

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(10)
YES