YES
proof of prog.inttrs
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given IRSwT could be proven:

(0) IRSwT
(1) IRSFormatTransformerProof [EQUIVALENT, 0 ms]
(2) IRSwT
(3) IRSwTTerminationDigraphProof [EQUIVALENT, 125 ms]
(4) IRSwT
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IRSwT
(7) TempFilterProof [SOUND, 50 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 15 ms]
(10) IntTRS
(11) RankingReductionPairProof [EQUIVALENT, 0 ms]
(12) YES


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(0)
Obligation:
Rules:
l0(xHAT0) -> l1(xHATpost) :|: xHATpost = 1 && xHAT0 <= 0
l0(x) -> l1(x1) :|: x1 = 1 + x && 1 <= x
l2(x2) -> l3(x3) :|: x2 = x3 && 4 <= x2
l2(x4) -> l0(x5) :|: x4 = x5 && 1 + x4 <= 4
l1(x6) -> l2(x7) :|: x6 = x7
l4(x8) -> l1(x9) :|: x10 = 5 && x9 = x9
l5(x11) -> l4(x12) :|: x11 = x12
Start term: l5(xHAT0)

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(1) IRSFormatTransformerProof (EQUIVALENT)
Reformatted IRS to match normalized format (transformed away non-linear left-hand sides, !=, / and %).
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(2)
Obligation:
Rules:
l0(xHAT0) -> l1(xHATpost) :|: xHATpost = 1 && xHAT0 <= 0
l0(x) -> l1(x1) :|: x1 = 1 + x && 1 <= x
l2(x2) -> l3(x3) :|: x2 = x3 && 4 <= x2
l2(x4) -> l0(x5) :|: x4 = x5 && 1 + x4 <= 4
l1(x6) -> l2(x7) :|: x6 = x7
l4(x8) -> l1(x9) :|: x10 = 5 && x9 = x9
l5(x11) -> l4(x12) :|: x11 = x12
Start term: l5(xHAT0)

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(3) IRSwTTerminationDigraphProof (EQUIVALENT)
Constructed termination digraph!
Nodes:
(1) l0(xHAT0) -> l1(xHATpost) :|: xHATpost = 1 && xHAT0 <= 0
(2) l0(x) -> l1(x1) :|: x1 = 1 + x && 1 <= x
(3) l2(x2) -> l3(x3) :|: x2 = x3 && 4 <= x2
(4) l2(x4) -> l0(x5) :|: x4 = x5 && 1 + x4 <= 4
(5) l1(x6) -> l2(x7) :|: x6 = x7
(6) l4(x8) -> l1(x9) :|: x10 = 5 && x9 = x9
(7) l5(x11) -> l4(x12) :|: x11 = x12

Arcs:
(1) -> (5)
(2) -> (5)
(4) -> (1), (2)
(5) -> (3), (4)
(6) -> (5)
(7) -> (6)

This digraph is fully evaluated!
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(4)
Obligation:

Termination digraph:
Nodes:
(1) l0(xHAT0) -> l1(xHATpost) :|: xHATpost = 1 && xHAT0 <= 0
(2) l2(x4) -> l0(x5) :|: x4 = x5 && 1 + x4 <= 4
(3) l1(x6) -> l2(x7) :|: x6 = x7
(4) l0(x) -> l1(x1) :|: x1 = 1 + x && 1 <= x

Arcs:
(1) -> (3)
(2) -> (1), (4)
(3) -> (2)
(4) -> (3)

This digraph is fully evaluated!

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
l2(x4:0) -> l2(1) :|: x4:0 < 4 && x4:0 < 1
l2(x) -> l2(1 + x) :|: x < 4 && x > 0

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(7) TempFilterProof (SOUND)
Used the following sort dictionary for filtering: 
l2(VARIABLE)
Replaced non-predefined constructor symbols by 0.
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(8)
Obligation:
Rules:
l2(x4:0) -> l2(c) :|: c = 1 && (x4:0 < 4 && x4:0 < 1)
l2(x) -> l2(c1) :|: c1 = 1 + x && (x < 4 && x > 0)

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[l2(x)] = -x

The following rules are decreasing:
l2(x4:0) -> l2(c) :|: c = 1 && (x4:0 < 4 && x4:0 < 1)
l2(x) -> l2(c1) :|: c1 = 1 + x && (x < 4 && x > 0)
The following rules are bounded:
l2(x4:0) -> l2(c) :|: c = 1 && (x4:0 < 4 && x4:0 < 1)

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(10)
Obligation:
Rules:
l2(x) -> l2(c1) :|: c1 = 1 + x && (x < 4 && x > 0)

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(11) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ l2 ] = -1*l2_1

The following rules are decreasing:
l2(x) -> l2(c1) :|: c1 = 1 + x && (x < 4 && x > 0)

The following rules are bounded:
l2(x) -> l2(c1) :|: c1 = 1 + x && (x < 4 && x > 0)


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(12)
YES