NO
proof of prog.inttrs
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given IRSwT could be disproven:

(0) IRSwT
(1) IRSFormatTransformerProof [EQUIVALENT, 0 ms]
(2) IRSwT
(3) IRSwTTerminationDigraphProof [EQUIVALENT, 62 ms]
(4) IRSwT
(5) IntTRSCompressionProof [EQUIVALENT, 0 ms]
(6) IRSwT
(7) FilterProof [EQUIVALENT, 3 ms]
(8) IntTRS
(9) IntTRSNonPeriodicNontermProof [COMPLETE, 0 ms]
(10) NO


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(0)
Obligation:
Rules:
l0 -> l1 :|: 0 <= 0
l1 -> l0 :|: 0 <= 0
l0 -> l2 :|: 0 <= 0
l2 -> l0 :|: 0 <= 0
l3 -> l2 :|: 0 <= 0
l4 -> l3 :|: 0 <= 0
Start term: l4

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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 -> l1 :|: 0 <= 0
l1 -> l0 :|: 0 <= 0
l0 -> l2 :|: 0 <= 0
l2 -> l0 :|: 0 <= 0
l3 -> l2 :|: 0 <= 0
l4 -> l3 :|: 0 <= 0
Start term: l4

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(3) IRSwTTerminationDigraphProof (EQUIVALENT)
Constructed termination digraph!
Nodes:
(1) l0 -> l1 :|: 0 <= 0
(2) l1 -> l0 :|: 0 <= 0
(3) l0 -> l2 :|: 0 <= 0
(4) l2 -> l0 :|: 0 <= 0
(5) l3 -> l2 :|: 0 <= 0
(6) l4 -> l3 :|: 0 <= 0

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

This digraph is fully evaluated!
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(4)
Obligation:

Termination digraph:
Nodes:
(1) l0 -> l1 :|: 0 <= 0
(2) l2 -> l0 :|: 0 <= 0
(3) l0 -> l2 :|: 0 <= 0
(4) l1 -> l0 :|: 0 <= 0

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

This digraph is fully evaluated!

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(5) IntTRSCompressionProof (EQUIVALENT)
Compressed rules.
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(6)
Obligation:
Rules:
l0 -> l0 :|: TRUE

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(7) FilterProof (EQUIVALENT)
Used the following sort dictionary for filtering: 
l0()
Replaced non-predefined constructor symbols by 0.
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(8)
Obligation:
Rules:
l0 -> l0 :|: TRUE

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(9) IntTRSNonPeriodicNontermProof (COMPLETE)
Normalized system to the following form:
f(pc) -> f(1) :|: pc = 1 && TRUE
Proved unsatisfiability of the following formula, indicating that the system is never left after entering:
((run2_0 = ((1 * 1)) and (((run1_0 * 1)) = ((1 * 1)) and T)) and !(((run2_0 * 1)) = ((1 * 1)) and T))
Proved satisfiability of the following formula, indicating that the system is entered at least once:
(run2_0 = ((1 * 1)) and (((run1_0 * 1)) = ((1 * 1)) and T))

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(10)
NO