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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 74 ms]
(4) IRSwT
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IRSwT
(7) TempFilterProof [SOUND, 15 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 0 ms]
(10) YES


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(0)
Obligation:
Rules:
l0(xHAT0) -> l1(xHATpost) :|: xHATpost = -1 + xHAT0 && 1 <= xHAT0
l1(x) -> l0(x1) :|: x = x1
l2(x2) -> l0(x3) :|: x2 = x3
Start term: l2(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 && 1 <= xHAT0
l1(x) -> l0(x1) :|: x = x1
l2(x2) -> l0(x3) :|: x2 = x3
Start term: l2(xHAT0)

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(3) IRSwTTerminationDigraphProof (EQUIVALENT)
Constructed termination digraph!
Nodes:
(1) l0(xHAT0) -> l1(xHATpost) :|: xHATpost = -1 + xHAT0 && 1 <= xHAT0
(2) l1(x) -> l0(x1) :|: x = x1
(3) l2(x2) -> l0(x3) :|: x2 = x3

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

This digraph is fully evaluated!
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(4)
Obligation:

Termination digraph:
Nodes:
(1) l0(xHAT0) -> l1(xHATpost) :|: xHATpost = -1 + xHAT0 && 1 <= xHAT0
(2) l1(x) -> l0(x1) :|: x = x1

Arcs:
(1) -> (2)
(2) -> (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) -> l0(-1 + xHAT0:0) :|: xHAT0:0 > 0

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

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[l0(x)] = x

The following rules are decreasing:
l0(xHAT0:0) -> l0(c) :|: c = -1 + xHAT0:0 && xHAT0:0 > 0
The following rules are bounded:
l0(xHAT0:0) -> l0(c) :|: c = -1 + xHAT0:0 && xHAT0:0 > 0

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(10)
YES