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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 75 ms]
(4) IntTRS
(5) IntTRSCompressionProof [EQUIVALENT, 26 ms]
(6) IntTRS
(7) PolynomialOrderProcessor [EQUIVALENT, 5 ms]
(8) AND
    (9) IntTRS
        (10) TerminationGraphProcessor [EQUIVALENT, 9 ms]
        (11) IntTRS
        (12) RankingReductionPairProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) IntTRS
        (15) IntTRSCompressionProof [EQUIVALENT, 0 ms]
        (16) IntTRS
        (17) RankingReductionPairProof [EQUIVALENT, 7 ms]
        (18) 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
f5(x4, x5) -> f8(arith, x5) :|: TRUE && arith = x4 - 1
f9(x32, x33) -> f12(x34, x33) :|: TRUE && x34 = x32 - 1
f10(x35, x36) -> f13(x35, x37) :|: TRUE && x37 = x36 - 1
f6(x10, x11) -> f9(x10, x11) :|: x10 = x11
f6(x12, x13) -> f10(x12, x13) :|: x12 < x13
f6(x38, x39) -> f10(x38, x39) :|: x38 > x39
f12(x14, x15) -> f11(x14, x15) :|: TRUE
f13(x16, x17) -> f11(x16, x17) :|: TRUE
f4(x18, x19) -> f5(x18, x19) :|: x18 > x19
f4(x20, x21) -> f6(x20, x21) :|: x20 <= x21
f8(x22, x23) -> f7(x22, x23) :|: TRUE
f11(x24, x25) -> f7(x24, x25) :|: TRUE
f3(x26, x27) -> f4(x26, x27) :|: x26 + x27 > 0
f7(x28, x29) -> f3(x28, x29) :|: TRUE
f3(x30, x31) -> f14(x30, x31) :|: x30 + x31 <= 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(x26, x27) -> f4(x26, x27) :|: x26 + x27 > 0
f7(x28, x29) -> f3(x28, x29) :|: TRUE
f8(x22, x23) -> f7(x22, x23) :|: TRUE
f5(x4, x5) -> f8(arith, x5) :|: TRUE && arith = x4 - 1
f4(x18, x19) -> f5(x18, x19) :|: x18 > x19
f11(x24, x25) -> f7(x24, x25) :|: TRUE
f12(x14, x15) -> f11(x14, x15) :|: TRUE
f9(x32, x33) -> f12(x34, x33) :|: TRUE && x34 = x32 - 1
f6(x10, x11) -> f9(x10, x11) :|: x10 = x11
f4(x20, x21) -> f6(x20, x21) :|: x20 <= x21
f13(x16, x17) -> f11(x16, x17) :|: TRUE
f10(x35, x36) -> f13(x35, x37) :|: TRUE && x37 = x36 - 1
f6(x12, x13) -> f10(x12, x13) :|: x12 < x13
f6(x38, x39) -> f10(x38, x39) :|: x38 > x39

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
f7(x28:0, x28:01) -> f7(x28:0 - 1, x28:0) :|: x28:0 + x28:0 > 0 && x28:0 = x28:01
f7(x, x1) -> f7(x, x1 - 1) :|: x + x1 > 0 && x1 > x
f7(x2, x3) -> f7(x2 - 1, x3) :|: x2 + x3 > 0 && x3 < x2

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(7) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[f7(x, x1)] = -1 + x1

The following rules are decreasing:
f7(x, x1) -> f7(x, x1 - 1) :|: x + x1 > 0 && x1 > x
The following rules are bounded:
f7(x28:0, x28:01) -> f7(x28:0 - 1, x28:0) :|: x28:0 + x28:0 > 0 && x28:0 = x28:01

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(8)
Complex Obligation (AND)

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(9)
Obligation:
Rules:
f7(x28:0, x28:01) -> f7(x28:0 - 1, x28:0) :|: x28:0 + x28:0 > 0 && x28:0 = x28:01
f7(x2, x3) -> f7(x2 - 1, x3) :|: x2 + x3 > 0 && x3 < x2

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

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(11)
Obligation:
Rules:
f7(x2, x3) -> f7(x2 - 1, x3) :|: x2 + x3 > 0 && x3 < x2

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(12) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f7 ] = f7_1 + -1*f7_2

The following rules are decreasing:
f7(x2, x3) -> f7(x2 - 1, x3) :|: x2 + x3 > 0 && x3 < x2

The following rules are bounded:
f7(x2, x3) -> f7(x2 - 1, x3) :|: x2 + x3 > 0 && x3 < x2


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(13)
YES

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(14)
Obligation:
Rules:
f7(x, x1) -> f7(x, x1 - 1) :|: x + x1 > 0 && x1 > x
f7(x2, x3) -> f7(x2 - 1, x3) :|: x2 + x3 > 0 && x3 < x2

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(15) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(16)
Obligation:
Rules:
f7(x:0, x1:0) -> f7(x:0, x1:0 - 1) :|: x:0 + x1:0 > 0 && x:0 < x1:0
f7(x2:0, x3:0) -> f7(x2:0 - 1, x3:0) :|: x2:0 + x3:0 > 0 && x3:0 < x2:0

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(17) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ f7 ] = f7_1 + f7_2

The following rules are decreasing:
f7(x:0, x1:0) -> f7(x:0, x1:0 - 1) :|: x:0 + x1:0 > 0 && x:0 < x1:0
f7(x2:0, x3:0) -> f7(x2:0 - 1, x3:0) :|: x2:0 + x3:0 > 0 && x3:0 < x2:0

The following rules are bounded:
f7(x:0, x1:0) -> f7(x:0, x1:0 - 1) :|: x:0 + x1:0 > 0 && x:0 < x1:0
f7(x2:0, x3:0) -> f7(x2:0 - 1, x3:0) :|: x2:0 + x3:0 > 0 && x3:0 < x2:0


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(18)
YES