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Modus tollens

Derive a denied antecedent from a conditional and a denied consequent.

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.

1. What you will learn

You will apply modus tollens to simple and compound formulas, write its proof references, and distinguish it from denying the antecedent.

2. Follow the arrow carefully

Modus ponens uses an antecedent to reach a consequent. Now start from a denial of the consequent and ask what that rules out.

3. Terms for the inference

4. Why modus tollens works

Suppose P -> Q and ~Q are true. Q is false. If P were true, the conditional would have a true antecedent and false consequent, which is its one false case. Therefore P cannot be true; ~P follows. The argument does not reverse the original conditional. It uses the denial of the required result to rule out its sufficient condition.

Compare denying the antecedent: P -> Q, ~P, therefore ~Q. The assignment P false and Q true makes its premises true and conclusion false. A result may have another sufficient condition. Ruling out one route does not rule out the result itself.

Treat compound formulas as whole units. From (P & Q) -> R and ~R, MT yields ~(P & Q). It does not give ~P & ~Q. At least one part of the conjunction must fail, but the premises do not identify which one. P true and Q false is compatible with the denied conjunction.

A denied consequent can itself be a double negation. From P -> ~Q and ~~Q, MT yields ~P. From P -> Q and Q you cannot use MT, because Q affirms rather than denies the consequent. Check the exact shape before choosing a named rule.

Denials can travel through a chain. From P -> Q, Q -> R and ~R, first infer ~Q by MT using the second conditional, then infer ~P using the first. Each inference cites its own conditional and matching denial. Writing both operations as one unexplained jump hides which rule produced the intermediate result.

5. Necessary consequences and ruled-out conditions

A conditional P -> Q presents Q as necessary for P within the stated claim: P cannot hold without Q. This is the useful perspective for modus tollens. If Q is ruled out, P is ruled out as well. You have not learned that Q would have been sufficient for P. You have learned that P requires something the second premise denies.

Take a fictional equipment guarantee: if the motor runs, the power indicator is on. If the indicator is not on, the guarantee and that observation jointly rule out a running motor. The inference depends on the guarantee being stated without an exception relevant to this case. If a failed indicator can stay dark while the motor runs, the simple conditional does not accurately represent the equipment. A valid use of MT cannot repair a false description of the system.

The same distinction appears in arguments with no causal interpretation. If a shape is a square, it has four sides. A shape without four sides cannot be a square. The inference concerns a necessary feature, not an event that caused another event. Modus tollens works from the logical relationship whether its subject is a definition, a stipulated rule, or an empirical regularity accepted as a premise. The evidence needed to justify those different sorts of premises is not identical.

To check an MT argument by a countermodel search, try making its conclusion ~P false. That makes P true. The conditional then requires Q true, but the second premise ~Q requires Q false. The requirements conflict, so no assignment can make both premises true while falsifying the conclusion. This short search supplies the semantic justification for the rule.

Contrast an argument that starts with ~P instead of ~Q. Setting P false and Q true satisfies P -> Q and ~P, yet falsifies ~Q. The conditional permits Q to hold in the absence of P. This single assignment explains why denying an antecedent is not a safe substitute for denying a consequent. The error is about which possibility the arrow excludes, not about whether the speaker chose a respectable example.

6. Negation reaches a formula, not a convenient part

Before applying MT, put brackets around the complete antecedent and the complete consequent. In (P & Q) -> (R | S), those parts are P & Q and R | S. A premise ~(R | S) denies the whole consequent, so MT yields ~(P & Q). Merely knowing ~R would not be enough: S could still make the disjunction true. Likewise, the conclusion does not establish ~P individually or ~Q individually. It denies their conjunction.

You can test an overstrong proposed conclusion by keeping one conjunct true and the other false. P true and Q false make ~(P & Q) true, but make ~P false. Therefore replacing the legitimate result by ~P rules out a case that the legitimate result permits. This is a useful correction strategy when a proof checker rejects a line that seems close to the answer: test the difference in scope, rather than swapping symbols at random.

An antecedent can already be a negation. From ~P -> Q and ~Q, MT derives ~~P. The rule denies the complete antecedent ~P, producing a double negation. In the classical two-valued logic used here, DN permits replacing ~~P by P on a further line. Write both operations if the allowed rules require them. Treating the MT result as P immediately would conceal the double-negation step and make the written reference check less precise.

Similarly, if the consequent is ~Q, its denial is ~~Q. A line Q can be converted to ~~Q with DN when that rule is allowed, then used with P -> ~Q in MT to derive ~P. The rule still requires a denial of the actual consequent, not a denial of the letter that happens to occur inside it. These examples are valuable because they prevent identifying inference rules by the mere presence of a negation sign.

The proof editor lists the rules available for a particular task. Knowing that another classical rule is sound does not make it available in a restricted exercise. If DN is omitted, give the exact negated formula that MT produces rather than silently simplifying it. Restrictions of this kind make an exercise assess the individual step being taught and let a reader inspect every transformation.

7. Tracing a denial and checking the evidence for it

For a chain P -> Q and Q -> R, a denial of R first applies to the second conditional. It produces ~Q. That derived denial is then the input required by the first conditional, producing ~P. The denial travels against the direction of the arrows, but each individual operation is justified by its own matching conditional and denied consequent. There is no general permission to reverse every arrow in an argument.

Write the intermediate ~Q even when the overall result looks immediate. A proof with line numbers makes it possible to identify exactly which stated rule carries the denial from R to Q and which carries it from Q to P. If one conditional has a different antecedent, the chain may stop. For example, P -> Q and S -> R do not connect P to R. From ~R you may infer ~S, but nothing in those premises then denies Q.

An absence of evidence is not automatically the evidence of absence that an MT premise needs. A missing entry in a complete, up-to-date record may support a denial under the record's stated completeness assumptions. A failed search, an inaccessible file, or an unexamined record may support only uncertainty. If the premise is 'we do not know Q', writing ~Q changes the claim. The logic lesson therefore asks you to distinguish an asserted denial from a report about somebody's information.

When using an actual negative observation, ask what would make that observation trustworthy. Was the relevant condition checked at the right time? Could the method miss it? Is the guarantee intended to hold without the exception now under discussion? These questions do not add new inference rules; they concern whether the premises supplied to MT are justified. Keeping the questions separate preserves both the exact logical result and the practical uncertainty that may remain.

The final proof audit has a short target: identify the complete consequent, locate its exact denial, negate the complete antecedent, and check the cited line numbers. Then state the result with its proper scope. A bounded conclusion such as 'these premises rule out both conditions holding together' can be fully established even when they do not tell you which individual condition failed.

8. A failed confirmation check

A fictional booking process guarantees that an accepted booking has a confirmation record. Let A mean the booking is accepted and C mean its confirmation record exists. The premises are A -> C and ~C. The conclusion ~A follows by MT. If A were true, the rule would force C, conflicting with the supplied absence of C.

This application relies on the complete stated guarantee and on an actual record absence. A search screen that has not loaded is not evidence of ~C; it may represent missing information. Likewise, if the process sometimes accepts bookings before writing confirmations, A -> C is not the right premise for that time. The logical inference cannot fix inaccurate modeling or incomplete evidence.

Contrast a booking known not to be accepted. A -> C and ~A do not imply ~C: perhaps a provisional record was created earlier. The assignment A false and C true satisfies these premises and refutes that conclusion. The two inferences can sound similar in ordinary language, which is why writing the arrow and its precise denial is useful.

For a longer workflow, suppose acceptance guarantees scheduling and scheduling guarantees confirmation. Given the absence of a confirmation, infer absence of scheduling first, then absence of acceptance. That is two MT steps, not one rule with three unrelated inputs. State the intermediate claim and cite the lines supporting it so that an auditor can check each dependency.

9. Negate the whole antecedent

If the antecedent is P | Q, its denial is ~(P | Q). If it is P & Q, its denial is ~(P & Q). Parentheses keep the scope visible. MT does not permit selecting an arbitrary part to deny. A useful check is to substitute the proposed values back into the conditional and the denial. If your inferred claim rules out an assignment that still satisfies both premises, you have concluded too much. For example, a denied conjunction still allows one conjunct to be true. Keep that possibility open unless another premise rules it out.

10. Extract a denial before using it

1. Identify the exact target.

   ~P

   Planning starts from the complete target formula, not one part of it.

2. Copy proof line 1.

   1. ~Q & R (premise)

   This formula is supplied by the problem and needs no inference or reference.

3. Copy proof line 2.

   1. P -> Q (premise)

   This formula is supplied by the problem and needs no inference or reference.

4. Write proof line 3.

   1. ~Q (andE 1)

   A true conjunction requires each of its parts to be true.

5. Write proof line 4.

   1. ~P (MT 2,3)

   A true antecedent would force the consequent that the other cited line denies.

11. Follow a denial through a chain

1. Identify the exact target.

   ~P

   Planning starts from the complete target formula, not one part of it.

2. Copy proof line 1.

   1. P -> Q (premise)

   This formula is supplied by the problem and needs no inference or reference.

3. Copy proof line 2.

   1. Q -> R (premise)

   This formula is supplied by the problem and needs no inference or reference.

4. Copy proof line 3.

   1. ~R (premise)

   This formula is supplied by the problem and needs no inference or reference.

5. Write proof line 4.

   1. ~Q (MT 2,3)

   A true antecedent would force the consequent that the other cited line denies.

6. Write proof line 5.

   1. ~P (MT 1,4)

   A true antecedent would force the consequent that the other cited line denies.

12. A negative antecedent produces a double denial

1. Identify the exact target.

   P

   Planning starts from the complete target formula, not one part of it.

2. Copy proof line 1.

   1. ~P -> Q (premise)

   This formula is supplied by the problem and needs no inference or reference.

3. Copy proof line 2.

   1. Q -> R (premise)

   This formula is supplied by the problem and needs no inference or reference.

4. Copy proof line 3.

   1. ~R (premise)

   This formula is supplied by the problem and needs no inference or reference.

5. Write proof line 4.

   1. ~Q (MT 2,3)

   A true antecedent would force the consequent that the other cited line denies.

6. Write proof line 5.

   1. ~~P (MT 1,4)

   A true antecedent would force the consequent that the other cited line denies.

7. Write proof line 6.

   1. P (DN 5)

   In classical two-valued logic, denying a denial restores the original truth value.

13. A compound antecedent

1. Copy the conditional and denial.

   (P & Q) -> R; ~R

   The denial matches the consequent R.

2. Negate the whole antecedent.

   ~(P & Q)

   MT denies the complete condition before the arrow.

3. Your turn: work this step out. Its working is at the end of the packet.

   Check what remains possible.

14. Guided practice

From P -> Q and ~Q derive ~P.

P -> Q
~Q
∴ ~P

15. Guided practice

Complete a worked audit of a denial traveling through P -> Q and Q -> R, with ~R supplied. Count derived denials, not premises.

1. Apply MT to Q -> R and ~R.

   ~Q; derived denials so far = _{after_first}

   Q would require the R that is ruled out.

2. Apply MT to P -> Q and the new ~Q.

   ~P; derived denials so far = _{after_second}

   P would require the Q that the previous inference ruled out.

3. Check the final scope.

   ~P

   The conclusion denies P; it does not reverse either conditional.

16. Guided practice

From P -> Q, Q -> R and ~R derive ~P.

P -> Q
Q -> R
~R
∴ ~P

17. Practice

Test the inference P -> Q, ~P, therefore ~Q. Supply a countermodel if it is invalid.

P -> Q
~P
∴ ~Q

valid invalid — countermodel:

18. Practice

From P -> (Q | R) and ~(Q | R), derive ~P.

P -> (Q | R)
~(Q | R)
∴ ~P

19. Somewhere new

A means accepted and C means confirmation exists. Given A -> C and ~C, derive ~A.

A -> C
~C
∴ ~A

20. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

21. Test question

From (P & Q) -> R and ~R derive ~(P & Q).

(P & Q) -> R
~R
∴ ~(P & Q)

22. What you can do now

You can identify exactly which consequent is denied and negate the whole matching antecedent.

Working for the steps left to you

13. A compound antecedent, step 3