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Working backwards from a false conclusion to the one row that refutes an argument, and why a failed search that closes every route is a proof of validity.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to construct a countermodel by falsifying the conclusion and satisfying the premises, fill in the whole table of an argument and read the breaking row off it, count the countermodels an argument has, refute an argument over three atoms without writing eight rows, and assign a value to an atom that occurs only in the conclusion.
You know that an argument is invalid exactly when some valuation makes every premise true and the conclusion false. This lesson is about producing that valuation rather than waiting for a table to reveal it.
| Term | What it means |
|---|---|
| Countermodel | An assignment satisfying all premises and falsifying the conclusion. |
| Constraint | A required truth value or combination imposed during the search. |
| Branch | One possible way to satisfy a compound requirement. |
| Exhaustion | Checking every remaining possibility, rather than abandoning a difficult search. |
A table finds a countermodel by exhaustion; with three atoms that is eight rows and with five it is thirty-two, so the practical method runs the other way. Start from what a countermodel has to do and work backwards through the connectives. Make the conclusion false: if it is a conditional, that forces the antecedent true and the consequent false; if it is a conjunction, one conjunct must fail and there is a choice; if it is a disjunction, both must fail and there is none. Then take the premises one at a time and see whether they can still be satisfied with the atoms that are already fixed. If every route closes with a contradiction, the argument is valid, and the closed routes are the proof. If a route stays open, fill in the atoms it left free, in any way at all, and you have a countermodel. One is enough; there is nothing to gain by finding a second.
Another way: steps
Another way: example
From $P \to Q$ and $\neg P$, conclude $\neg Q$. The conclusion is false only with $Q$ true; the second premise needs $P$ false; the first is then satisfied. Countermodel: $P$ false, $Q$ true.
Suppose a conclusion is P | Q. To make it false, both atoms must be false; there is only one possibility for those two values. If instead the conclusion is P & Q, making it false has three possibilities. Trying just P false and Q false is insufficient: a premise might force P true while allowing Q false.
For premises P and P -> (Q | R), conclusion Q, start with Q false. P must be true. The conditional then requires Q | R true; with Q false, set R true. The complete assignment P true, Q false, R true satisfies the premises and refutes the conclusion. The search succeeds because we let the disjunction's other branch do the work.
Always finish by substituting the proposed values back into every original premise. A value chosen early in the search may conflict with a later premise. If all possibilities close, explain which constraint closes each branch or use an exhaustive table. 'I could not find one' describes an unfinished search; 'all assignments fail one required condition' establishes that no countermodel exists.
A countermodel is not just a situation in which the conclusion is false. It is a situation in which the conclusion is false despite every stated premise being true. These two requirements belong together. If your story changes a premise, omits a condition, or switches the meaning of a letter, it may challenge something else but does not refute the specified inference.
Write the complete premise list and target before assigning values. Mark each premise with the requirement true and the conclusion with the requirement false. Then work inward through the connectives to determine atomic constraints. This makes the search directed: you are constructing the one kind of assignment that would defeat entailment rather than trying unrelated rows and hoping one works.
An atomic conclusion is straightforward to falsify: set that atom false. A negated atomic conclusion requires the atom true. A conditional conclusion requires true antecedent and false consequent. A conjunction conclusion permits multiple ways of being false, while a disjunction conclusion requires both sides false. Those differences control how many branches the search must preserve.
Suppose the premises are P | Q and P -> R, with conclusion R. Set R false. To keep P -> R true, set P false. The disjunction can still be true if Q is true. The assignment P false, Q true, R false satisfies both premises and falsifies R. The alternative disjunct was essential: assuming P true would have produced a conflict and might have led you to abandon the search too early.
Once the values are chosen, evaluate the original formulas again from scratch. Do not verify only a simplified note made during the search. The original list may contain a condition you forgot to propagate. A final audit showing every premise value and the conclusion value is the certificate that turns a candidate into a verified countermodel.
A branching requirement represents genuine alternatives, not uncertainty about the truth rules. To make P & Q false, you can have P false and Q true, P true and Q false, or both false. A premise might eliminate two of these while leaving the third. Testing just the both-false assignment would then miss a valid countermodel.
For the argument P; therefore P & Q, the premise forces P true. The both-false candidate fails the premise, but P true and Q false works. This illustrates why a failed first attempt does not establish validity. Diagnose why the attempt failed, retain the constraint P true, and try a conclusion-false branch compatible with that constraint.
Disjunction in a premise creates another kind of choice. If P | Q must be true, either true disjunct can support it, and both may be true. Other premises can force one branch. With ~P added, Q must be true. Without ~P, setting P true may be a perfectly good option. Record which restriction closes a branch rather than treating your first convenient choice as logically forced.
Sometimes an atom remains free after the decisive constraints are satisfied. For premises P and Q | R with conclusion Q, set Q false, then P true and R true. Every atom is now fixed. In a different argument with an extra unused atom S, either value of S could extend that countermodel. Give a complete assignment if the interface asks for all displayed atoms, but do not pretend the unconstrained value was required by the reasoning.
To prove that no countermodel exists, show that every branch closes or use a complete table. A branch closes when it demands incompatible values for one atom or directly falsifies a required premise. A search that simply feels difficult remains unfinished. The distinction protects valid arguments from premature rejection and invalid arguments from premature acceptance.
One verified countermodel proves the argument invalid under the declared propositional representation. It does not prove that the conclusion is actually false, that the speaker is dishonest, or that all related arguments fail. An invalid inference can accidentally reach a true conclusion. The countermodel shows only that the stated premises do not guarantee it.
Turning the assignment into a concrete story can help explain the missing link. In the library example, a guest admitted without a pass realizes pass false and entry permitted true. The point is not to prove that such a guest actually visited. It is to show a possibility the stated rule permits. If a further rule excludes guests, add it explicitly and analyze the strengthened premise set instead of silently assuming it was present all along.
A background restriction can legitimately change the result, but it must be part of the modeled information. If a system uses an exclusive choice, a countermodel with both alternatives true may violate that extra restriction. If the prompt declares inclusive or, the same assignment is allowed. Read the modeling conventions carefully and distinguish an unstated expectation from a supplied premise.
Countermodels also help improve an argument constructively. After identifying a surviving assignment, ask which missing premise would exclude it and whether there is independent evidence for that premise. Adding the desired conclusion as a premise would block all countermodels, but it would not supply an informative defense of the conclusion to somebody who doubts it. A useful repair explains the missing relationship rather than merely assuming the result.
When several countermodels exist, choose one that makes the failure easy to inspect. A compact assignment with a clear concrete interpretation is often more helpful than a long list of equivalent failures. Still include all required atomic values and every premise check. The strongest explanatory feature is not the number of examples but the transparent preservation of all premises while the target fails.
Finally, keep the result tied to the exact argument tested. Changing an arrow's direction, replacing inclusive or with exclusive or, or adding a premise produces a different argument. An old countermodel may stop working after such a change. Re-evaluate it against the revised formulas rather than carrying over the invalidity label by resemblance.
A library's fictional rule says: if a visitor has a pass, the visitor may enter. A visitor may enter. Someone concludes that the visitor has a pass. Let P mean pass held and E mean entry permitted. The premises are P -> E and E, and the conclusion is P. To refute the inference, make P false. The second premise already requires E true. On P false and E true, the conditional is true, the second premise is true, and the conclusion is false. This is a complete countermodel.
A concrete story matching the assignment is a visitor admitted as a guest without a pass. The story is not needed for the table calculation, but it helps explain the missing restriction: the rule makes a pass sufficient for entry, without making it necessary.
Now consider a proposed countermodel with P false and E false. It also makes the conclusion false, but it fails the premise that entry is permitted. It therefore does not refute this argument. The two candidates differ in exactly the value that checks whether the counterexample respects the given evidence.
One valid countermodel is enough; counting more does not make the inference more invalid. Conversely, an actual visitor with a pass does not rescue the inference. The issue is whether the stated premises leave open a passless visitor who may enter. If the library intends passes to be necessary as well, it must add E -> P, then test the strengthened argument separately.
The first error is offering a row that makes the conclusion false without checking the premises on it; such a row refutes nothing, because the argument never promised anything there. The second is leaving an atom unassigned — most often one that appears only in the conclusion — which leaves a half-built valuation rather than a countermodel. The third is concluding validity from a failed search that was never exhaustive: the search must close every route, not merely the first two tried.
State the argument.
P; therefore P & Q
The target claims more than the atomic premise.
Keep the premise true.
P=T
A countermodel must preserve P.
Make the other conjunct false.
Q=F
No premise requires Q true.
Evaluate the target.
P & Q=T & F=F
One false conjunct falsifies the conjunction.
Verify and conclude.
P=T,Q=F; invalid
The premise is true while the conclusion is false.
Record the target and premises.
P | Q, P -> R; therefore R
Both premises must survive the search.
Falsify the conclusion.
R=F
This is required for a countermodel.
Preserve the conditional.
P=F
With R false, a true P would violate the arrow.
Preserve the disjunction.
Q=T
Q must supply the true disjunct once P is false.
Audit the original formulas.
P | Q=T; P -> R=T; R=F
All checks use P=F,Q=T,R=F.
State the countermodel.
P=F,Q=T,R=F; invalid
The alternate branch leaves the conclusion unguaranteed.
State the argument.
P | Q, ~R; therefore P & Q
A false conjunction has several possible assignments.
Try both conjuncts false.
P=F,Q=F
This certainly falsifies the conclusion.
Check the first premise.
P | Q=F
The attempted row fails a premise and is not a countermodel.
Retain a different false-conclusion branch.
P=T,Q=F
The conjunction remains false while the disjunction becomes true.
Satisfy the second premise.
R=F
This makes ~R true.
Audit the completed assignment.
P | Q=T; ~R=T; P & Q=F
Every original requirement is checked on P=T,Q=F,R=F.
Conclude despite the failed first attempt.
Invalid
A later verified branch is sufficient; the initial failure did not exhaust the search.
State the candidate for P -> Q,Q; therefore P.
P=F,Q=F
The conclusion P is false.
Check the atomic premise.
Q=F
The candidate fails a required premise.
Repair the assignment.
Construct a countermodel if possible for P -> (Q | R), Q; therefore P. If none exists, decide validity from a complete constraint check.
P -> (Q | R)
Q
∴ P
valid invalid — countermodel:
Audit candidate P=F,Q=F for premises P -> Q and Q, conclusion P. Count true premises, then change Q to T and recount.
Evaluate the first candidate.
P -> Q=T; Q=F; before true premise
A false conclusion alone does not make this a countermodel.
Evaluate the repaired candidate.
P=F,Q=T; after true premises
Both the arrow and the atomic assertion now hold.
Inspect the conclusion again.
P=F; invalid
The repaired assignment preserves every premise and falsifies the target.
Construct a countermodel if possible for P | Q; therefore P -> Q. If none exists, decide validity from a complete constraint check.
P | Q
∴ P -> Q
valid invalid — countermodel:
Construct a countermodel if possible for P -> Q, ~P; therefore ~Q. If none exists, decide validity from a complete constraint check.
P -> Q
~P
∴ ~Q
valid invalid — countermodel:
P means a pass is held and E means entry is permitted. Express that holding a pass is sufficient for entry, without saying it is necessary.
Answer:
Construct a countermodel if possible for P | Q, P -> R, Q -> R; therefore R. If none exists, decide validity from a complete constraint check.
P | Q
P -> R
Q -> R
∴ R
valid invalid — countermodel:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Construct a countermodel if possible for (P & Q) -> R, ~R; therefore ~P. If none exists, decide validity from a complete constraint check.
(P & Q) -> R
~R
∴ ~P
valid invalid — countermodel:
You can produce a countermodel for an invalid argument and explain why a search that closes every route establishes validity. Say in your own words why a row that makes the conclusion false is not yet a countermodel.
14. An invalid proposed countermodel, step 3
P=F,Q=T
Now both premises hold while P remains false.