/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 42 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 2 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(8) IntTRS
(9) RankingReductionPairProof [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
f3(x1, x2) -> f4(arith, x2) :|: TRUE && arith = x1 + 1
f4(x3, x4) -> f5(x3, 1) :|: TRUE
f6(x21, x22) -> f7(x21, x23) :|: TRUE && x23 = x22 + 1
f5(x7, x8) -> f6(x7, x8) :|: x7 >= x8
f7(x9, x10) -> f5(x9, x10) :|: TRUE
f5(x11, x12) -> f8(x11, x12) :|: x11 < x12
f8(x24, x25) -> f9(x26, x25) :|: TRUE && x26 = x24 - 2
f2(x15, x16) -> f3(x15, x16) :|: x15 >= 0
f9(x17, x18) -> f2(x17, x18) :|: TRUE
f2(x19, x20) -> f10(x19, x20) :|: x19 < 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:
f2(x15, x16) -> f3(x15, x16) :|: x15 >= 0
f9(x17, x18) -> f2(x17, x18) :|: TRUE
f8(x24, x25) -> f9(x26, x25) :|: TRUE && x26 = x24 - 2
f5(x11, x12) -> f8(x11, x12) :|: x11 < x12
f4(x3, x4) -> f5(x3, 1) :|: TRUE
f3(x1, x2) -> f4(arith, x2) :|: TRUE && arith = x1 + 1
f7(x9, x10) -> f5(x9, x10) :|: TRUE
f6(x21, x22) -> f7(x21, x23) :|: TRUE && x23 = x22 + 1
f5(x7, x8) -> f6(x7, x8) :|: x7 >= x8

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f5(x7:0, x8:0) -> f5(x7:0, x8:0 + 1) :|: x8:0 <= x7:0
f5(x11:0, x12:0) -> f5(x11:0 - 1, 1) :|: x12:0 > x11:0 && x11:0 > 1

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

The following rules are decreasing:
f5(x11:0, x12:0) -> f5(x11:0 - 1, 1) :|: x12:0 > x11:0 && x11:0 > 1
The following rules are bounded:
f5(x11:0, x12:0) -> f5(x11:0 - 1, 1) :|: x12:0 > x11:0 && x11:0 > 1

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(8)
Obligation:
Rules:
f5(x7:0, x8:0) -> f5(x7:0, x8:0 + 1) :|: x8:0 <= x7:0

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(9) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f5 ] = -1*f5_2 + f5_1

The following rules are decreasing:
f5(x7:0, x8:0) -> f5(x7:0, x8:0 + 1) :|: x8:0 <= x7:0

The following rules are bounded:
f5(x7:0, x8:0) -> f5(x7:0, x8:0 + 1) :|: x8:0 <= x7:0


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(10)
YES