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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 139 ms]
(4) IRSwT
(5) IntTRSCompressionProof [EQUIVALENT, 39 ms]
(6) IRSwT
(7) TempFilterProof [SOUND, 58 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(10) IntTRS
(11) RankingReductionPairProof [EQUIVALENT, 0 ms]
(12) YES


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(0)
Obligation:
Rules:
l0(xHAT0, yHAT0, zHAT0) -> l1(xHATpost, yHATpost, zHATpost) :|: zHAT0 = zHATpost && yHAT0 = yHATpost && xHAT0 = xHATpost && 1 <= yHAT0 && 1 <= xHAT0
l1(x, x1, x2) -> l0(x3, x4, x5) :|: x2 = x5 && x3 = x2 && x4 = -1 + x1
l1(x6, x7, x8) -> l0(x9, x10, x11) :|: x8 = x11 && x7 = x10 && x9 = -1 + x6
l2(x12, x13, x14) -> l0(x15, x16, x17) :|: x14 = x17 && x13 = x16 && x12 = x15
l3(x18, x19, x20) -> l2(x21, x22, x23) :|: x20 = x23 && x19 = x22 && x18 = x21
Start term: l3(xHAT0, yHAT0, zHAT0)

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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, yHAT0, zHAT0) -> l1(xHATpost, yHATpost, zHATpost) :|: zHAT0 = zHATpost && yHAT0 = yHATpost && xHAT0 = xHATpost && 1 <= yHAT0 && 1 <= xHAT0
l1(x, x1, x2) -> l0(x3, x4, x5) :|: x2 = x5 && x3 = x2 && x4 = -1 + x1
l1(x6, x7, x8) -> l0(x9, x10, x11) :|: x8 = x11 && x7 = x10 && x9 = -1 + x6
l2(x12, x13, x14) -> l0(x15, x16, x17) :|: x14 = x17 && x13 = x16 && x12 = x15
l3(x18, x19, x20) -> l2(x21, x22, x23) :|: x20 = x23 && x19 = x22 && x18 = x21
Start term: l3(xHAT0, yHAT0, zHAT0)

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(3) IRSwTTerminationDigraphProof (EQUIVALENT)
Constructed termination digraph!
Nodes:
(1) l0(xHAT0, yHAT0, zHAT0) -> l1(xHATpost, yHATpost, zHATpost) :|: zHAT0 = zHATpost && yHAT0 = yHATpost && xHAT0 = xHATpost && 1 <= yHAT0 && 1 <= xHAT0
(2) l1(x, x1, x2) -> l0(x3, x4, x5) :|: x2 = x5 && x3 = x2 && x4 = -1 + x1
(3) l1(x6, x7, x8) -> l0(x9, x10, x11) :|: x8 = x11 && x7 = x10 && x9 = -1 + x6
(4) l2(x12, x13, x14) -> l0(x15, x16, x17) :|: x14 = x17 && x13 = x16 && x12 = x15
(5) l3(x18, x19, x20) -> l2(x21, x22, x23) :|: x20 = x23 && x19 = x22 && x18 = x21

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

This digraph is fully evaluated!
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(4)
Obligation:

Termination digraph:
Nodes:
(1) l0(xHAT0, yHAT0, zHAT0) -> l1(xHATpost, yHATpost, zHATpost) :|: zHAT0 = zHATpost && yHAT0 = yHATpost && xHAT0 = xHATpost && 1 <= yHAT0 && 1 <= xHAT0
(2) l1(x6, x7, x8) -> l0(x9, x10, x11) :|: x8 = x11 && x7 = x10 && x9 = -1 + x6
(3) l1(x, x1, x2) -> l0(x3, x4, x5) :|: x2 = x5 && x3 = x2 && x4 = -1 + x1

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

This digraph is fully evaluated!

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
l0(xHAT0:0, yHAT0:0, x3:0) -> l0(x3:0, -1 + yHAT0:0, x3:0) :|: xHAT0:0 > 0 && yHAT0:0 > 0
l0(x, x1, x2) -> l0(-1 + x, x1, x2) :|: x > 0 && x1 > 0

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(7) TempFilterProof (SOUND)
Used the following sort dictionary for filtering: 
l0(VARIABLE, INTEGER, VARIABLE)
Replaced non-predefined constructor symbols by 0.
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(8)
Obligation:
Rules:
l0(xHAT0:0, yHAT0:0, x3:0) -> l0(x3:0, c, x3:0) :|: c = -1 + yHAT0:0 && (xHAT0:0 > 0 && yHAT0:0 > 0)
l0(x, x1, x2) -> l0(c1, x1, x2) :|: c1 = -1 + x && (x > 0 && x1 > 0)

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[l0(x, x1, x2)] = -1 + x1

The following rules are decreasing:
l0(xHAT0:0, yHAT0:0, x3:0) -> l0(x3:0, c, x3:0) :|: c = -1 + yHAT0:0 && (xHAT0:0 > 0 && yHAT0:0 > 0)
The following rules are bounded:
l0(xHAT0:0, yHAT0:0, x3:0) -> l0(x3:0, c, x3:0) :|: c = -1 + yHAT0:0 && (xHAT0:0 > 0 && yHAT0:0 > 0)
l0(x, x1, x2) -> l0(c1, x1, x2) :|: c1 = -1 + x && (x > 0 && x1 > 0)

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(10)
Obligation:
Rules:
l0(x, x1, x2) -> l0(c1, x1, x2) :|: c1 = -1 + x && (x > 0 && x1 > 0)

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(11) RankingReductionPairProof (EQUIVALENT)
Interpretation:
[ l0 ] = l0_1

The following rules are decreasing:
l0(x, x1, x2) -> l0(c1, x1, x2) :|: c1 = -1 + x && (x > 0 && x1 > 0)

The following rules are bounded:
l0(x, x1, x2) -> l0(c1, x1, x2) :|: c1 = -1 + x && (x > 0 && x1 > 0)


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(12)
YES