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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) TRUE


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(0)
Obligation:
Rules:
l0(aHAT0, bHAT0) -> l1(aHATpost, bHATpost) :|: bHAT0 = bHATpost && aHAT0 = aHATpost && 1 <= 0
l1(x, x1) -> l0(x2, x3) :|: x1 = x3 && x = x2
l0(x4, x5) -> l2(x6, x7) :|: x4 = x6 && x7 = 0
l2(x8, x9) -> l3(x10, x11) :|: x9 = x11 && x8 = x10 && 1 + x9 <= 1
l4(x12, x13) -> l0(x14, x15) :|: x13 = x15 && x12 = x14 && 1 + x12 <= 1
l4(x16, x17) -> l2(x18, x19) :|: x16 = x18 && x19 = 1
l5(x20, x21) -> l4(x22, x23) :|: x21 = x23 && x22 = 1
l6(x24, x25) -> l5(x26, x27) :|: x25 = x27 && x24 = x26
Start term: l6(aHAT0, bHAT0)

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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(aHAT0, bHAT0) -> l1(aHATpost, bHATpost) :|: bHAT0 = bHATpost && aHAT0 = aHATpost && 1 <= 0
l1(x, x1) -> l0(x2, x3) :|: x1 = x3 && x = x2
l0(x4, x5) -> l2(x6, x7) :|: x4 = x6 && x7 = 0
l2(x8, x9) -> l3(x10, x11) :|: x9 = x11 && x8 = x10 && 1 + x9 <= 1
l4(x12, x13) -> l0(x14, x15) :|: x13 = x15 && x12 = x14 && 1 + x12 <= 1
l4(x16, x17) -> l2(x18, x19) :|: x16 = x18 && x19 = 1
l5(x20, x21) -> l4(x22, x23) :|: x21 = x23 && x22 = 1
l6(x24, x25) -> l5(x26, x27) :|: x25 = x27 && x24 = x26
Start term: l6(aHAT0, bHAT0)

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(3) IRSwTTerminationDigraphProof (EQUIVALENT)
Constructed termination digraph!
Nodes:
(1) l0(aHAT0, bHAT0) -> l1(aHATpost, bHATpost) :|: bHAT0 = bHATpost && aHAT0 = aHATpost && 1 <= 0
(2) l1(x, x1) -> l0(x2, x3) :|: x1 = x3 && x = x2
(3) l0(x4, x5) -> l2(x6, x7) :|: x4 = x6 && x7 = 0
(4) l2(x8, x9) -> l3(x10, x11) :|: x9 = x11 && x8 = x10 && 1 + x9 <= 1
(5) l4(x12, x13) -> l0(x14, x15) :|: x13 = x15 && x12 = x14 && 1 + x12 <= 1
(6) l4(x16, x17) -> l2(x18, x19) :|: x16 = x18 && x19 = 1
(7) l5(x20, x21) -> l4(x22, x23) :|: x21 = x23 && x22 = 1
(8) l6(x24, x25) -> l5(x26, x27) :|: x25 = x27 && x24 = x26

Arcs:
(2) -> (3)
(3) -> (4)
(5) -> (3)
(7) -> (6)
(8) -> (7)

This digraph is fully evaluated!
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(4)
TRUE