Disjunction and the chain rule

Disjunctive syllogism, the free-but-useless disjunction introduction, the chain rule for conditionals, and the two biconditional rules.

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

By the end of this lesson you will be able to use disjunctive syllogism, disjunction introduction, the chain rule and the biconditional rules in a derivation, say what each demands of the lines it cites, explain why a freely added disjunct can never be recovered, and choose a first move when no rule fits a premise as it stands.

2. What you already have

You can write a derivation with modus ponens, modus tollens and the two conjunction rules. This lesson adds the rules for disjunction, the rule that joins two conditionals, and the two rules for the biconditional.

3. The words this lesson uses

An introduction rule builds a connective into a line; an elimination rule takes one out. DS is disjunctive syllogism, orI disjunction introduction, HS hypothetical syllogism or the chain rule, and biE and biI the two rules for the biconditional.

4. Disjunction and the chain rule

DS: from $\phi \vee \psi$ and $\neg \phi$, write $\psi$ — a disjunction needs at least one disjunct, so denying one leaves the other. orI: from $\phi$ alone, write $\phi \vee \psi$ for any $\psi$. That looks like getting something for nothing, and it is the opposite: the new line is weaker, true on every row where $\phi$ is and on more besides, so nothing about $\psi$ has been claimed and nothing about it can be recovered. HS: from $\phi \to \psi$ and $\psi \to \chi$, write $\phi \to \chi$ — the shared letter disappears, and the rule applies again to its own output, so a chain of any length collapses to its two ends. biE: from $\phi \leftrightarrow \psi$, write either conditional; biI: from both conditionals, write the biconditional. A premise no rule fits is not a dead end but a sign that the first move is an elimination.

Another way: steps

Look at the shape of each premise and ask which rules could touch it.
Eliminate a connective that blocks every rule — a biconditional, a conjunction.
For a disjunction, look for a line denying one disjunct.
For two conditionals, check whether they share a letter in the right positions.

Another way: example

From $P \vee Q$, $\neg P$ and $Q \to R$: DS on the first two gives $Q$, and modus ponens then gives $R$. Three premises, two steps.

5. Where this goes wrong

The first error is using disjunctive syllogism with an affirmed disjunct rather than a denied one, which is the fallacy of affirming a disjunct. The second is expecting to get back what disjunction introduction added: the rule claimed nothing about it, so nothing about it follows. The third is chaining two conditionals that share an antecedent rather than a consequent and antecedent — $\phi \to \psi$ and $\phi \to \chi$ chain into nothing at all.

6. A denial that opens a disjunction

Lines 1 and 2: $P \vee Q$ and $\neg Q$.
A disjunction and a denial.
The denial closes the second disjunct, so line 3 is $P$ by DS.
The other disjunct survives.
Had line 2 been $Q$ rather than $\neg Q$, nothing would follow: both disjuncts may hold at once.
Denied, not affirmed.

7. Chaining twice

$P \to Q$, $Q \to R$, $R \to S$. The first two share $Q$.
Consequent meets antecedent.
Line 4: $P \to R$ by HS. That is a conditional, so it can chain again.
The output is reusable.
Line 5: $P \to S$ by HS from line 4 and the third premise.
Two joins, three premises.

8. Your turn: from $P \leftrightarrow Q$ and $P$, derive $Q$

No rule for conditionals touches a biconditional, so eliminate it first: line 3 is $P \to Q$ by biE.
Your turn: work this step out. Its working is at the end of the packet.
Line 4 is $Q$, by modus ponens from lines 3 and 2.

9. Guided practice

From the premises P | Q, ~P and Q -> R, derive R. Give one line at a time, with the rule and the lines it uses.

P | Q
~P
Q -> R
∴ R

#	Formula	Rule	Lines
1
2
3
4
5
6

10. Guided practice

From the premises Q -> R, R -> S and S -> P, derive Q -> P. Give one line at a time, with the rule and the lines it uses.

Q -> R
R -> S
S -> P
∴ Q -> P

#	Formula	Rule	Lines
1
2
3
4
5
6

11. Guided practice

Match each rule to what it demands and what it lets you write.

	Cite a disjunction and the denial of one disjunct; write the other disjunct	Cite any line; write it disjoined with any formula whatever	Cite two conditionals that share a letter; write the conditional joining the two ends	Cite a biconditional; write either of the conditionals it contains
Disjunctive syllogism
Disjunction introduction
Hypothetical syllogism
Biconditional elimination

12. Practice

Here is an attempted proof. Mark every line that the rule it names does not produce.

This task has no paper form; do it on a device.

13. Practice

Disjunction introduction lets you write $\phi \vee \psi$ from $\phi$ alone, whatever $\psi$ is. What makes that legitimate?

Because a disjunction is true on every row where either half is, so the new line is true wherever the cited line is
Because the formula being added must be true as well
Because the two halves of a disjunction may be interchanged
Because the added disjunct can be removed again by disjunctive syllogism

14. Somewhere new

From the premises P <-> Q and Q -> R, derive P -> R. Give one line at a time, with the rule and the lines it uses.

P <-> Q
Q -> R
∴ P -> R

#	Formula	Rule	Lines
1
2
3
4
5
6

15. Lesson test

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

16. Test question

These are the five lines of a proof, shuffled. Put them in order.

Number the steps in order (write the number in the box):

1. $P \to Q$, premise
2. $Q \to R$, premise
3. $\neg R$, premise
4. $P \to R$, from 1 and 2 by hypothetical syllogism
5. $\neg P$, from 4 and 3 by modus tollens

17. What you can do now

You can build a derivation using the disjunction, chaining and biconditional rules, and find the lines of somebody else's that their rules do not produce. Say in your own words why disjunction introduction is safe and yet gains you nothing about the formula it adds.

Working for the steps left to you

8. Your turn: from $P \leftrightarrow Q$ and $P$, derive $Q$, step 2