Hypothetical and Disjunctive Syllogism (HS, DS)

Two more rules of inference come from the valid forms in Valid and Invalid Argument Forms: hypothetical syllogism (HS), which chains conditionals, and disjunctive syllogism (DS), which eliminates one side of an "or". They build on Modus Ponens and Modus Tollens (MP, MT).

Hypothetical and disjunctive syllogism.
Rule	From	Infer
HS	$p \to q$ and $q \to r$	$p \to r$
DS	$p \lor q$ and $\neg p$	$q$
DS	$p \lor q$ and $\neg q$	$p$

Key idea

HS: two conditionals that share a middle part give the conditional from the first antecedent to the last consequent. DS: a disjunction and the negation of one disjunct give the other disjunct.

Why it works

HS is valid because if $p$ leads to $q$ and $q$ leads to $r$, then $p$ leads to $r$: a counterexample would need $p$ T and $r$ F, and then $q$ could be neither T (breaking $q \to r$) nor F (breaking $p \to q$). DS is valid because a true disjunction has at least one true disjunct; if one is false, the other must be the true one.

DS, then HS, then MP.

Worked examples

Worked example: Chain conditionals

From $\neg p \to r$ and $r \to (s \land t)$, what does HS give?

The consequent of the first, $r$, is the antecedent of the second.
HS joins the outer parts.

Answer: $\neg p \to (s \land t)$

Worked example: Disjunctive syllogism

From $(p \land q) \lor r$ and $\neg (p \land q)$, what does DS give?

Line 2 is the negation of the first disjunct.
DS gives the other disjunct.

Answer: $r$

Worked example: Order of the lines

From $q \to r$ on line 1 and $p \to q$ on line 2, can you use HS?

HS does not care which line comes first.
Line 2 leads into line 1: $p$ to $q$, $q$ to $r$.

Answer: Yes: $p \to r$, justified "HS 2, 1" (or "HS 1, 2").

Common mistake

Using DS with an affirmed disjunct: from $p \lor q$ and $p$, inferring $\neg q$. Because "or" is inclusive, both disjuncts may be true; knowing one is true tells you nothing about the other. DS needs the negation of one disjunct. (In Traditional Logic this mistake is called affirming a disjunct.)

DS needs the exact negation. From $\neg p \lor q$ and $p$ you cannot use DS yet, because $p$ is not written as $\neg \neg p$. After the rules of replacement, DN will fix that in one step.

Notice that both rules have two versions or two orders. DS may eliminate either disjunct, the left or the right, and HS may cite its two conditionals in either order. In each case the shapes must still line up: HS needs the middle formula to be exactly the same in both conditionals.

HS and DS often meet in the same proof: DS frees one part of a disjunction, and HS or MP carries it the rest of the way to the conclusion.

Hypothetical and Disjunctive Syllogism (HS, DS)

Worked examples

Practice

Builds on

Leads to

Sources and further reading