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Assuming a formula, deriving a contradiction and discharging the assumption as a negation, with double negation for the last step and modus tollens for when no assumption is needed.
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 lay out a proof by reductio, identify the two lines that form its contradiction, say what a derived contradiction does and does not show, reach a negation by modus tollens when no assumption is needed, test an argument by adding the denial of its conclusion, and use double negation to bridge a formula and its double negation.
You can assume a formula, work under the assumption and discharge it. Reductio is the same machinery with a different target: instead of deriving a consequent, you derive a contradiction, and what is written at the discharge is a negation.
A contradiction is a pair of lines, one the negation of the other, or the single line $\phi \wedge \neg \phi$. Reductio ad absurdum assumes a formula, derives a contradiction and discharges the assumption as its negation. Double negation is the rule taking $\neg \neg \phi$ to $\phi$ and back.
To prove $\neg \phi$: assume $\phi$, derive a contradiction from it together with the premises, and discharge, writing $\neg \phi$. The reasoning is the same as conditional proof's. A contradiction is true on no row, so no row satisfies everything the derivation used; the premises were granted and the assumption was only supposed, so the assumption is what gives way. It follows that a reductio says nothing whatever about the premises — they may be perfectly consistent, and it is their combination with $\phi$ that is not. To prove $\phi$ rather than $\neg \phi$, assume $\neg \phi$ instead; the discharge then gives $\neg \neg \phi$, and double negation takes the last step. Reductio is never the only route: when modus tollens or the chain rule reaches a negation directly, that is shorter, and reductio is what remains when they do not.
Another way: steps
Another way: example
Premises $P \to Q$ and $P \to \neg Q$; goal $\neg P$. Assume $P$; both premises fire, giving $Q$ and $\neg Q$; discharge to get $\neg P$. The premises themselves are consistent — both hold whenever $P$ is false.
The first error is concluding that the premises are inconsistent: the contradiction was derived from the premises and the assumption, and only the assumption was supposed. The second is keeping a line derived under the assumption after the discharge, which carries a supposition out of the subproof. The third is treating $\neg \neg \phi$ as though it were already $\phi$: they are equivalent, so the step is available, but it is a step and a rule has to take it.
Premises $P \to Q$ and $\neg Q$; goal $\neg P$. Assume $P$.
Assume what you mean to refute.
Modus ponens gives $Q$, and the second premise is $\neg Q$: a contradiction.
Two lines, one the negation of the other.
Discharge: $\neg P$. Modus tollens would have given the same line in one step, and here it was available.
Shorter routes first.
Premises $P \vee Q$ and $\neg Q$; goal $P$. Assume $\neg P$.
Assume the denial of the goal.
Disjunctive syllogism gives $Q$, against the second premise.
Contradiction again.
Discharge gives $\neg \neg P$, and double negation gives $P$.
One extra step at the end.
Assume $P$. Both premises are conditionals with $P$ as antecedent, so both fire.
They give $Q$ and $\neg Q$; discharging the assumption gives $\neg P$, and the premises remain consistent with each other.
These are the six lines of a proof by reductio, shuffled. Put them in order.
Number the steps in order (write the number in the box):
Here is a proof by reductio. Mark the two lines that form the contradiction.
This task has no paper form; do it on a device.
From the premises ~R, Q -> R and P -> Q, derive ~P. Give one line at a time, with the rule and the lines it uses.
~R
Q -> R
P -> Q
∴ ~P
| # | Formula | Rule | Lines |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 6 |
You assumed $P$ and, using the premises, derived both $Q$ and $\neg Q$. What have you shown?
Match each rule to what it demands and what it lets you write.
| Assume a formula, derive a contradiction, write its negation | Cite a doubly negated line; write the line itself | Cite a conditional and the denial of its consequent; write the denial of its antecedent | Cite two lines; write their conjunction, which is how a contradiction is displayed | |
|---|---|---|---|---|
| Reductio ad absurdum | ||||
| Double negation elimination | ||||
| Modus tollens | ||||
| Conjunction introduction |
From the premises ~~P -> Q and P, derive Q. Give one line at a time, with the rule and the lines it uses.
~~P -> Q
P
∴ Q
| # | Formula | Rule | Lines |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 6 |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Premises: P | Q and ~Q. Conclusion: P. Does the conclusion follow? If it does not, give a row that breaks it.
P | Q
~Q
∴ P
valid invalid — countermodel:
You can build a reductio and say exactly what it establishes. Say in your own words why a derived contradiction blames the assumption rather than the premises.
8. Your turn: from $P \to Q$ and $P \to \neg Q$, prove $\neg P$, step 2