Tutorial 10

Many-valued logics

Exercise sheet 10 questions

K₃ SAT solving

Use SQL to decide whether each inference is valid in K₃. The table Valuations(SUN, RAIN, WIND) is already filled with all 3³ = 27 combinations of 0, NULL, and 1. As in the chapter, NULL represents ω.

Write a query that returns all three columns for exactly the countermodels: each premise IS TRUE, while the conclusion IS NOT TRUE. Run it, then choose ⊨ or ⊭ to interpret the result.

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Solution

For the first inference, we search for countermodels like this:

1
2
3

SELECT SUN, RAIN, WIND FROM Valuations
WHERE (RAIN OR (RAIN AND SUN)) IS TRUE
  AND SUN IS NOT TRUE;

Levels 1, 2, and 5 are invalid. Countermodels include, respectively, v(RAIN) = 1, v(SUN) = 0; v(SUN) = 1, v(RAIN) = 0; and v(WIND) = 1, v(RAIN) = ω. The unspecified values can be chosen freely. Levels 3, 4, and 6 are valid: their queries return no rows. For level 3, no valuation makes both premises true. For level 5, notice that an unknown conclusion is enough for a countermodel.

K₃ derivations

Use the K₃ rules to derive each conclusion from the displayed premises. The negation levels include double negations inside compound formulas; the case arguments combine De Morgan with disjunction elimination. Identify which assumptions are temporary and discharge them. Write material conditionals as ¬A ∨ B.

Explosion takes two premises, A and ¬A, in either order, and lets us infer any C. There is no need to conjoin the premises first.

Solution

K₃ laws

Verify the Kleene laws one at a time. Each level displays an equation; write a query returning exactly the rows where its two sides disagree. The table ValueTriples(X, Y, Z) already contains all 27 combinations of 0, NULL, and 1. Return all three columns, even when the equation uses fewer variables.

Use IS NOT to compare the two sides. Ordinary != returns NULL if either side is NULL, and WHERE would discard that row. The last level tests a proposed identity that fails: your query should find its counterexamples.

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Solution

For associativity of AND, for example:

1
2

SELECT X, Y, Z FROM ValueTriples
WHERE (X AND (Y AND Z)) IS NOT ((X AND Y) AND Z);

All the Kleene laws return no rows. For the last level, use WHERE (X OR NOT X) IS NOT 1. This returns the nine rows with X = NULL: the result does not depend on Y or Z.

Łukasiewicz conditionals

The chapter’s fuzzy conditional has value min(1, 1 − v(A) + v(B)). As in the chapter, write 0.5 for the intermediate value ω. Restricting the conditional to 0, 0.5, and 1 gives this table:

→00.51
0111
0.50.511
100.51

This gives us three-valued Łukasiewicz logic. Here, as in K₃, validity means preserving the designated value 1. Later in these exercises, we return to the stronger, degree-preserving notion of fuzzy consequence.

  1. Find a K₃ model where the material conditional RAIN → RAIN is not true.
  2. Verify that RAIN → RAIN has value 1 in every Łukasiewicz model.
  3. Is modus ponens valid in three-valued Łukasiewicz logic?

Solution

  1. If v(RAIN) = ω, the K₃ material conditional has value max(1 − 0.5, 0.5) = 0.5, so it is not true.
  2. Whatever v(RAIN) is, min(1, 1 − v(RAIN) + v(RAIN)) = 1.
  3. Yes. If v(A) = 1, then v(A → B) = min(1, v(B)) = v(B). Thus, if both premises have value 1, so does B.

Fuzzy predicates

Let’s interpret Fast¹ and Slow¹ for speeds from 0 to 120 km/h. In every level, require that Fast never decreases as speed increases, while Slow never increases. At 0 km/h the car is fully slow and not fast; at 120 km/h it is fully fast and not slow.

Select a predicate and pull its curve up or down. Dots mark the speeds mentioned in the task. You can also choose a speed and enter its degree. The levels compare overlapping descriptions with a gap between them.

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Solution

For the complement levels, choose the degree of Fast at each point and set Slow to 1 − Fast. An overlapping gray area can have both degrees 0.5 throughout 40–80 km/h.

For the gap level, let Slow reach 0 by 40 km/h and let Fast remain 0 until 80 km/h. They cannot be complements in that interval: both descriptions fail to apply.

Fuzzy rules

Choose interpretations that make the displayed rules hold at every speed. Here, only the rules and any stated degree bounds constrain your choices. Recall that A ⇒ B requires the degree of A to be at most the degree of B. After completing the levels, explain how the two rules together relate Fast and Slow.

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Solution

Fast(x) ⇒ ¬Slow(x) requires Fast(x) + Slow(x) ≤ 1. The reverse rule requires Fast(x) + Slow(x) ≥ 1. Together they force Fast(x) + Slow(x) = 1: the predicates are complements. Choosing complementary interpretations also solves the first two levels, although neither rule by itself requires complementation. For the last level, we can keep both degrees at 0.5 throughout the gray band.

Fuzzy counterexamples

Find a counterexample at each level. For degree-preserving consequence, the conclusion must have at least the least degree of the premises. To refute the inference, make its value strictly smaller. With no premises, the required value is 1.

Here → is the Łukasiewicz conditional from the chapter: v(A → B) = min(1, 1 − v(A) + v(B)). Compare the modus ponens level with the three-valued exercise above: why do the two questions have different answers?

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Solution

For modus ponens, choose v(A) = 0.6 and v(B) = 0.3. The conditional has value 0.7, but min(0.6, 0.7) = 0.6 > 0.3. This does not refute preservation of 1: the premises aren’t both 1.

For both disjunctive syllogism and explosion, choose v(A) = 0.5 and v(B) = 0. The least premise degree is 0.5 and the conclusion has degree 0. For affirming the consequent, choose v(A) = 0, v(B) = 1; these same values refute denying the antecedent. Excluded middle fails at v(A) = 0.5.

Valid fuzzy inferences

Prove that the following inferences are valid for degree-preserving fuzzy consequence. Your calculation needs to work for arbitrary degrees a = v(A) and b = v(B). Compare the least premise degree with the conclusion’s degree using min and max.

Fuzzy inferences to prove

Conjunction elimination

Conjunction introduction

Disjunction introduction

Solution

Let a = v(A) and b = v(B) be any values in [0, 1]. For conjunction elimination, we have:

v(A ∧ B) = min(a, b) ≤ a = v(A).

For conjunction introduction, the least premise degree is exactly the conclusion’s degree:

min(v(A), v(B)) = min(a, b) = v(A ∧ B).

For disjunction introduction, we have:

v(A) = a ≤ max(a, b) = v(A ∨ B).

Each calculation holds for all a and b, as required.

Combining controller rules

A train has received a braking signal. While the signal is active, our controller uses the train’s speed to choose braking power. We want stronger braking at high speeds and gentler braking as the train slows down. Use these interpretations of Slow, Moderate, and Fast:

Slow, Moderate, and Fast
Degree 00.51 0406080120 Speed (km/h) ⟦Moderate⟧⟦Fast⟧⟦Slow⟧

These predicates overlap: at 60 km/h, the speed is fully Moderate and counts as both Slow and Fast to degree 0.5. Their degrees need not add up to 1.

Power runs from 0 (no braking) to 1 (full braking); 0.5 means half power. Use this scale throughout.

For Sugeno, choose the recommended outputs c₁, c₂, and c₃ for Slow(x) ↦ c₁, Moderate(x) ↦ c₂, and Fast(x) ↦ c₃. Their weights come from the diagram:

w₁ = ⟦Slow⟧(x), w₂ = ⟦Moderate⟧(x), w₃ = ⟦Fast⟧(x) power = (w₁c₁ + w₂c₂ + w₃c₃) / (w₁ + w₂ + w₃).

Meet the power requirements at each level. Level 3 also asks for a smooth join at 60 km/h: the curve’s direction should not change abruptly there. The dashed guides and the check help you find suitable outputs.

The final level uses Mamdani: choose the Low, Medium, and High output sets shown in the app. The Medium triangle extends 0.2 to either side of its peak.

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Explain in words how different settings can meet the same requirements. Which Sugeno output would you change to increase power at 60 km/h without changing it at 0 or 120 km/h? Why? What relationship between the three outputs removes the corner in level 3?

Solution

For the first level, choose (c₁, c₂, c₃) = (0.1, 0.5, 0.9). The powers at 0, 60, and 120 km/h are then 0.1, 0.5, and 0.9.

For the second, choose (c₁, c₂, c₃) = (0, 0.8, 1). At 60 km/h the weights are 0.5, 1, and 0.5, so:

power = (0.5×0 + 1×0.8 + 0.5×1) / (0.5 + 1 + 0.5) = 0.65.

At 0 and 120 km/h, only Slow or Fast applies, giving powers 0 and 1. The total weight changes from 1 at the endpoints to 2 at 60 km/h. Dividing by this changing total produces a curved response, even though the membership curves consist of straight segments.

For level 3, (c₁, c₂, c₃) = (0.1, 0.5, 0.9) works. With these interpretations, the Moderate output must be halfway between the other two:

c₂ = (c₁ + c₃) / 2.

For example, (0, 0.4, 0.8) also works. Raising or lowering c₂ from this midpoint introduces a corner. With the midpoint setting, the two parts of the curved response meet without an abrupt change in direction.

For Mamdani, let Low reach 0 at power 0.2, let Medium peak at 0.75, and let High start rising at 0.6. The powers are approximately 0.067, 0.625, and 0.867 at the three speeds.

There are many solutions because the tasks specify ranges. For example, (0, 0.5, 0.8) also solves the first level. Raising any recommendation increases power wherever its condition applies. Raising c₂ increases power at 60 km/h while leaving the endpoints unchanged, since Moderate has degree 0 at both endpoints. In Mamdani, we can move the centroid by changing the output sets’ shapes and positions.

What does the intermediate value mean?

Compare these four cases. What information is represented, and which of our models would you consider using?

  1. A database has no recorded age for little Jimmy.
  2. A tank is half full.
  3. The probability of rain tomorrow is 0.5.
  4. Two reports conflict: one says it is raining, the other says it isn’t.

Solution

  1. This is missing information. We could leave both an age claim and its negation unentailed in a classical KB, or use a SQL-style unknown value.
  2. This is a measurement. We can represent the fraction directly as 0.5. A fuzzy predicate such as “full” needs a chosen membership function; its degree needn’t equal the measured fraction.
  3. This is uncertainty about an event. A probability of 0.5 does not mean that the event will be half true.
  4. This is conflicting information. We might keep track of sources and revise a belief, or use a paraconsistent consequence relation. Calling K3’s ω “both” by itself doesn’t change its designated values or give us a suitable treatment of the conflict.