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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 77 ms]
(4) IRSwT
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IRSwT
(7) TempFilterProof [SOUND, 21 ms]
(8) IntTRS
(9) PolynomialOrderProcessor [EQUIVALENT, 6 ms]
(10) YES


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(0)
Obligation:
Rules:
l0(xHAT0) -> l1(xHATpost) :|: xHAT0 = xHATpost
l1(x) -> l2(x1) :|: 0 <= x && x2 = x && x1 = -1 + x2
l2(x3) -> l1(x4) :|: x3 = x4
l3(x5) -> l0(x6) :|: x5 = x6
Start term: l3(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) :|: xHAT0 = xHATpost
l1(x) -> l2(x1) :|: 0 <= x && x2 = x && x1 = -1 + x2
l2(x3) -> l1(x4) :|: x3 = x4
l3(x5) -> l0(x6) :|: x5 = x6
Start term: l3(xHAT0)

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(3) IRSwTTerminationDigraphProof (EQUIVALENT)
Constructed termination digraph!
Nodes:
(1) l0(xHAT0) -> l1(xHATpost) :|: xHAT0 = xHATpost
(2) l1(x) -> l2(x1) :|: 0 <= x && x2 = x && x1 = -1 + x2
(3) l2(x3) -> l1(x4) :|: x3 = x4
(4) l3(x5) -> l0(x6) :|: x5 = x6

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

This digraph is fully evaluated!
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(4)
Obligation:

Termination digraph:
Nodes:
(1) l1(x) -> l2(x1) :|: 0 <= x && x2 = x && x1 = -1 + x2
(2) l2(x3) -> l1(x4) :|: x3 = x4

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:
l1(x2:0) -> l1(-1 + x2:0) :|: x2:0 > -1

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(7) TempFilterProof (SOUND)
Used the following sort dictionary for filtering: 
l1(INTEGER)
Replaced non-predefined constructor symbols by 0.
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(8)
Obligation:
Rules:
l1(x2:0) -> l1(c) :|: c = -1 + x2:0 && x2:0 > -1

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(9) PolynomialOrderProcessor (EQUIVALENT)
Found the following polynomial interpretation:
[l1(x)] = x

The following rules are decreasing:
l1(x2:0) -> l1(c) :|: c = -1 + x2:0 && x2:0 > -1
The following rules are bounded:
l1(x2:0) -> l1(c) :|: c = -1 + x2:0 && x2:0 > -1

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(10)
YES