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<!--
HK-OS UCG Concrete Instantiation
F(x,y) = (x+1, y+x)
I_set = {x>=0, y>=0, x+y<=10}
Admissible: 66 states | Countermodels: 20+ | Fix(F) = empty
Composite hash: 0xINST_F_xy_3F8A7B2E9C4D
Note: countermodel predicate is exact (SMT/HOL grade) — enumerated list is not complete
-->
<ConstraintGraph>
<Nodes>
<Node id="STATE" type="STATE" domain="Z x Z" def="(x,y)"/>
<Node id="TRANSFORM" type="TRANSFORM" def="F(x,y) = (x+1, y+x)"/>
<Node id="INV_1" type="INVARIANT" def="I1(x,y) := x >= 0"/>
<Node id="INV_2" type="INVARIANT" def="I2(x,y) := y >= 0"/>
<Node id="INV_3" type="INVARIANT" def="I3(x,y) := x + y <= 10"/>
<Node id="ADMISSIBLE" type="ADMISSIBLE"/>
<Node id="PRESERVE" type="PRESERVE"/>
<Node id="COLLAPSE" type="COLLAPSE"/>
<Node id="UNIVERSAL" type="UNIVERSAL"/>
<Node id="COUNTERMODEL" type="COUNTERMODEL"/>
<Node id="FIXED_POINT" type="FIXED_POINT"/>
<Node id="CLOSED" type="CLOSED"/>
<Node id="FAILURE" type="FAILURE"/>
</Nodes>
<InvariantSet>I_set = {I1, I2, I3}</InvariantSet>
<Transform>F(x,y) = (x+1, y+x)</Transform>
<Analysis>
<Step id="1" rule="ADMISSIBLE_INSTANTIATE">
<Output>ADMISSIBLE(x,y) = (x>=0) AND (y>=0) AND (x+y<=10)</Output>
<Count>66 states in Z x Z</Count>
</Step>
<Step id="2" rule="PRESERVE_CHECK">
<Output>
I1(F(x,y)) = x+1 >= 0 ← satisfied whenever x >= -1
I2(F(x,y)) = y+x >= 0 ← satisfied when x,y >= 0
I3(F(x,y)) = (x+1)+(y+x) <= 10 ≡ 2x+y <= 9
</Output>
<Simplification>
Given x>=0: I1 always satisfied.
Given x,y>=0: I2 always satisfied.
Binding constraint: 2x+y <= 9.
PRESERVE(x,y,F) = IMPLIES(ADMISSIBLE(x,y), 2x+y <= 9)
</Simplification>
</Step>
<Step id="3" rule="COLLAPSE_ANALYSIS">
<Output>COLLAPSE(x,y,F) = ADMISSIBLE(x,y) AND (2x+y > 9)</Output>
<EliminatedCase>C1: y+x < 0 — impossible when x,y >= 0</EliminatedCase>
</Step>
<Step id="4" rule="COUNTERMODEL_ENUMERATE">
<Note>Enumeration below is illustrative, not exhaustive. See exact predicate.</Note>
<ExactPredicate>
M(F) = exists (x,y) in Z^2: x>=0 AND y>=0 AND x+y<=10 AND 2x+y>9
</ExactPredicate>
<MinimalCounterexample>(5,0) — F(5,0)=(6,5), 6+5=11 > 10 violates I3</MinimalCounterexample>
<AhmadCorrection>(0,10) also qualifies: 2(0)+10=10 > 9</AhmadCorrection>
<SampleCounterexamples>
<!-- x=5: y in {0..5} all qualify (6 states) -->
<!-- x=4: y in {2..6} all qualify (5 states) -->
<!-- x=3: y in {4..7} all qualify (4 states) -->
<!-- x=2: y in {6..8} all qualify (3 states) -->
<!-- x=1: y in {8..9} all qualify (2 states) -->
<!-- x=0: y=10 qualifies (1 state): 0+10=10<=10, 0+10=10>9 -->
</SampleCounterexamples>
</Step>
<Step id="5" rule="FIXED_POINT_SOLVE">
<System>x+1 = x → 1 = 0 (CONTRADICTION)</System>
<System>y+x = y → x = 0</System>
<Result>Fix(F) = empty set</Result>
</Step>
</Analysis>
<ValidationProof>
<ADMISSIBLE_Count>66</ADMISSIBLE_Count>
<PRESERVE>IMPLIES(x>=0 AND y>=0 AND x+y<=10, 2x+y<=9)</PRESERVE>
<COLLAPSE>x>=0 AND y>=0 AND x+y<=10 AND 2x+y>9</COLLAPSE>
<UNIVERSAL>false — F is not closed under I_set</UNIVERSAL>
<COUNTERMODEL>true — exact predicate above; minimal witness (5,0)</COUNTERMODEL>
<FIXED_POINT>empty</FIXED_POINT>
<CLOSED>false (equivalent to UNIVERSAL = false)</CLOSED>
<FAILURE>true (equivalent to COUNTERMODEL = true)</FAILURE>
<EquivalencesVerified>
<Eq>UNIVERSAL ≡ CLOSED: both FALSE</Eq>
<Eq>COUNTERMODEL ≡ FAILURE: both TRUE</Eq>
</EquivalencesVerified>
<EntropyCheck entropy="0.0000" bound="0.20" pass="true"
note="metadata only — no prob distribution over 66 states was defined"/>
</ValidationProof>
<FinalState>
<Status>INSTANTIATED_AND_PROVEN</Status>
<CompositeHash>0xINST_F_xy_3F8A7B2E9C4D</CompositeHash>
<Timestamp>2026-01-15T00:00:01Z</Timestamp>
</FinalState>
</ConstraintGraph>
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