Unit Propagation

Lecture 7, DSCI 220, 2026W1

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Where we left off

Conjunctive Normal Form

An AND of clauses. Every expression has one…

Expression	Law
\(\neg(p \rightarrow q) \vee r\)

~(p -> q) v r ₍p v q) v r IMP (p ^ ~q) v r DM, then DNEG (p v r) ^ (~q v r) DIST <- CNF, two clauses

Verified: all four rows agree on all eight assignments.

Back to Inference

Where we started: Given statements (premises), a conclusion, and ten(ish) rules. Pick a rule, pick what to apply it to, repeat until conclusion.

• \(p \rightarrow q\)
• \(p\)
• \(q\)

So why is that hard to hand to a machine?

________________________________________

With CNF we only need ELIM…

Ten is a small number, so the rule count is not the problem – push back if someone says it is. The cost is the SEARCH:

• at every step you choose a rule AND which of your facts to feed it, so the branching is the rules times the subsets, not just the rules;
• each rule wants a different shape, so every choice needs a pattern match before you even know whether it applies;
• GEN makes the space infinite on its own – from p you may write p v q for ANY q, so there are always infinitely many legal next moves;
• and nothing tells you which branch reaches the goal. You cannot tell a promising line from a dead one until you are down it.

That is what CNF buys. One shape, so no matching. One rule, so no choosing. And because resolution returns what it consumes, you can apply it blindly to every pair and stop when nothing new appears – no cleverness anywhere.

Second blank: unit clause. Wednesday is unit propagation – given (X) and (~X v Y), drop ~X and keep Y. That is DS, which they have had since Friday under the name ELIM. Do not say so now; let them find it Wednesday.

Proving by refuting

To show a conclusion follows, show that denying it is impossible.

Inference
the conclusion follows from the premises
premises \(\rightarrow\) conclusion is a __________
\(\neg(\)premises \(\rightarrow\) conclusion\()\) is a __________
premises \(\wedge\ \neg\)conclusion is __________

tautology / contradiction / unsatisfiable – three ways of saying one thing, and the last is the one a machine can hunt for.

So the recipe: negate the goal, add it to the premises, put everything in CNF, and look for the empty clause.

Negating the Whole Thing

Say we start with an argument like this:

\[\neg\big((A \wedge B \wedge C) \rightarrow D\big)\]

Expression	Law
\(\neg\big(\neg(A \wedge B \wedge C) \vee D\big)\)

So the premises survive untouched and only the conclusion flips.

~((A^BC) -> D) ₍(A^BC) v D) IMP (A^BC) ^ ~D DM (A^BC) ^ ~D DNEG (then ASS to drop the brackets)

THE POINT. This is why the recipe is “keep the premises, negate the goal” – it is not a convention, it falls out of the algebra. Negating an implication never touches the antecedent.

And look at the CNF: A ^ B ^ C ^ ~D is already a conjunction of FOUR UNIT CLAUSES, {A} {B} {C} {~D}. Every premise arrives as fuel. That is why these problems start so easily, and it is worth pointing at before the worksheet, where some sets deliberately do not.

One Rule, Repeated

You know \((X)\). You also have \((\neg X \vee Y)\).

\(\neg X\) is false, so it cannot be what makes that clause true. Strike it.

What is left must be true: ______

Y. And they have known this rule since Friday – it is ELIM on the Sheet, disjunctive syllogism. The only new thing is doing it relentlessly instead of choosing it.

This is unit propagation.

The Loop

Repeat until you cannot:

1. Find a clause with exactly one literal. It must be true. Circle it.
2. Cross out every clause that contains it — satisfied, gone.
3. In every clause containing its negation, strike out that literal only.
4. A clause that loses its last literal must be False–a contradiction.

Three ways to stop:

Stop	Means
you reach False	the clauses contradict — the conclusion follows!!
no clauses left	all variables assigned — conclusion does not follow
clauses remain, none a literal	nothing more is forced

Your Turn

Every group has a different set. Annotate the sheet.

Answer two questions:

• What can you conclude?
• How do you know you are finished?

13 sets, in activities/lecture-07_activity.qmd. Five reach BOTTOM, four run out, four stall – so all three endings are in the room, which is the debrief.

ANSWER KEY – units in forced order:

Set 1 D => -C => -B => -E CONTRADICTION conclusion FOLLOWS Set 2 -B => G => -C => -F CONTRADICTION conclusion FOLLOWS Set 3 B => E => -A => -F CONTRADICTION conclusion FOLLOWS Set 4 C => -D => -B => A CONTRADICTION conclusion FOLLOWS Set 5 -D => G => E => C CONTRADICTION conclusion FOLLOWS Set 6 -G => A => -B => -D RUNS OUT conclusion does not follow Set 7 G => -D => A => C RUNS OUT conclusion does not follow Set 8 F => -A => B => -E RUNS OUT conclusion does not follow Set 9 B => D => A => -F RUNS OUT conclusion does not follow Set 10 -A => C => D STALLS conclusion does not follow Set 11 -E => F => A STALLS conclusion does not follow Set 12 -G => E => F STALLS conclusion does not follow Set 13 C => A => E STALLS conclusion does not follow

Every sequence is forced, so a group that gets a different one has made an error. The usual slip is a 3- or 4-literal clause: it only becomes a unit once two or three of its literals are dead, and it is easy to treat it like the two-literal ones.

What Did You Get?

Your ending	What it proves
False
no clauses left
stalled

BOTTOM: the premises plus the negated goal cannot all hold, so the argument is valid. Proved.

NO CLAUSES LEFT: satisfiable, so there IS a way to make the premises true and the conclusion false. Invalid – and the units they derived ARE the counterexample. Verified: in all four such sets the derived units satisfy every clause, whichever way the one free variable goes. Ask a group to read theirs out; it is a witness, not just a failure.

STALLED: proves nothing either way. All four stalling sets are satisfiable – two assignments each – so there was something to find and propagation could not find it.

That asymmetry is the point of the whole lecture. Reaching bottom proves validity. NOT reaching it proves nothing, unless you also ran out of clauses.

When Nothing Is Forced

\[(A\vee B)\ \wedge\ (A \vee \neg B)\ \wedge\ (\neg A \vee B)\ \wedge\ (\neg A \vee \neg B)\]

No clause has a single literal. The loop cannot even start.

And yet: is there any assignment that satisfies all four? ______

No – it is unsatisfiable, and all four rows of the table fail. So the answer is THERE, propagation just cannot reach it.

What would? Pick a variable and try it both ways. Assume A: clauses 3 and 4 give B and ~B. Assume ~A: clauses 1 and 2 give B and ~B. Both branches die, so it is unsatisfiable.

That case split is where a real SAT solver spends its life, and it is the honest limit of what we did today.

Words and Ideas

Refutation

Unit clause, unit propagation

The empty clause contradiction

Satisfiable, unsatisfiable