Modus Ponens

Context: FIT1058_MOC, FIT2014_MOC · the core rule of inference · turns $P$ and $P\Rightarrow Q$ into $Q$ · justifies each “logical consequence” step of a proof

Quick Revision

• 🎯 Objective: from $P$ and $P\Rightarrow Q$, deduce $Q$ ➔ the engine of every “logical consequence” proof step.
• 📦 Core Components: fact $P$ ➔ implication $P\Rightarrow Q$ ➔ conclusion $Q$.
• ⚡ Key Constraint: both premises required; deducing $P$ from $Q$ and $P\Rightarrow Q$ is the converse fallacy.

📝 Core

1. The Rule (Affirming the Antecedent)

• Statement ➔ if $P$ established and $P\Rightarrow Q$ established, then $Q$ deducible.
• Layout ➔ $\frac{P,P\Rightarrow Q}{\therefore Q}$.
• Role ➔ justifies each “logical consequence” step of a proof.

2. Usage & Limits

• Both needed ➔ $P$ alone or $P\Rightarrow Q$ alone is insufficient.
• Converse error ➔ from $Q$ and $P\Rightarrow Q$ you may not conclude $P$.
• Drives induction ➔ base $S(1)$ + step $S(k)\Rightarrow S(k+1)$ fires modus ponens repeatedly.

Key identities:

$P:X\in \mathcal{P}(A)(\text{true}),P\Rightarrow Q:\text{definition of }\mathcal{P}(A)$ $\therefore Q:X\subseteq A(\text{modus ponens})$

⚖️ Core Decision Matrix

Have	Rule	Conclude	Valid?
$P$ and $P\Rightarrow Q$	modus ponens	$Q$	✅
$Q$ and $P\Rightarrow Q$	(affirming consequent)	$P$	❌ converse fallacy
$\neg Q$ and $P\Rightarrow Q$	modus tollens	$\neg P$	✅

When It Flips: so natural it is rarely named in prose — "we have $P$, and $P$ implies $Q$, therefore $Q$" is just "a deduction". Its premises must themselves be established (definition/axiom/prior theorem/earlier step).

📊 Exam Execution Trace

Applied Exercise

Problem: Given "$6\mid 12$" and “if $6\mid n$ then $2\mid n$”, deduce a conclusion. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rlrl}P& :6\mid 12& & \text{(established fact)}\\ P\Rightarrow Q& :6\mid n\Rightarrow 2\mid n& & \text{(given implication)}\\ \therefore Q& :2\mid 12& & \text{(modus ponens)}\end{array}$$

Final Extracted Output: $2\mid 12$ by modus ponens; from $Q$ one may not conclude $P$ (converse fallacy).

⚠️ Common Mistakes

• 💡 Don’t affirm the consequent ➔ having $Q$ and $P\Rightarrow Q$ does not give $P$ (converse fallacy); you need the antecedent $P$ itself.

🧠 Active Recall

State modus ponens and explain why both premises are essential.

• Hint: Fact + conditional.

Answer

• Short answer: From $P$ and $P\Rightarrow Q$ deduce $Q$; the implication is inert without its antecedent holding.
• Why: Converse fallacy ➔ deducing $P$ from $Q$ and $P\Rightarrow Q$ is invalid.

How does modus ponens appear in a power-set subset proof?

• Hint: Definition as the implication.

Answer

• Short answer: With $P:X\in \mathcal{P}(A)$ and definitional $P\Rightarrow Q$ ($Q:X\subseteq A$), modus ponens yields $X\subseteq A$.
• Why: Membership = subset ➔ the Power Set definition supplies the conditional the rule consumes.