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MHT-CET Maths · Mathematical Logic

Logical Equivalence and Algebra of Statements

A short list of named laws lets you reach the answer to an equivalence question by algebra instead of by building the whole truth table.

All Mathematical Logic notesMHT-CET Maths strategy

Why this matters

34 PYQs sit here at 26% HARD, and this is the subtopic that decides how long the rest of the chapter takes you. You already know what equivalence MEANS — the first block defined it as two matching last columns. What is new here is the shortcut: an eight-row table always works but is slow and error-prone, while three lines of algebra reach the same verdict. The table is the method that always terminates; the algebra is the method that is fast. Build the table when the pattern is small, when the algebra stalls, or when the question hands you the truth values outright — reach for the laws when there are three letters and an arrow to clear. The same laws then do the work in Switching Circuits, so the time spent here is paid back twice.

Concept 1 of 6: Why Equivalence Licenses Substitution

You met A≡BA \equiv B in the first block as a reading of the last column: two patterns whose columns match row for row. What this block adds is the consequence — because they can never disagree, one may be SWAPPED for the other anywhere, and that permission is what makes every simplification below legal.

Definition

Recall A≡BA \equiv B: AA and BB have the same truth value in every row, equivalently A↔BA \leftrightarrow B is a tautology.

  • The substitution licence. Equivalent statements may replace each other inside any larger pattern without changing its truth value. Every law in the table below is used this way.
  • The asymmetry. To DISPROVE equivalence you need only ONE row where they differ; to ESTABLISH it you need every row, or an algebraic derivation that stands in for every row.
  • This is why the algebra exists at all: each law is a pre-proved equivalence you may apply without re-deriving its table.

Equivalence and its test

A≡B  ⟺  (A↔B) is a tautology  ⟺  A and B share every rowA \equiv B \iff (A \leftrightarrow B) \text{ is a tautology} \iff A \text{ and } B \text{ share every row}
  • ≡\equivlogically equivalent — a claim about all rows
  • ↔\leftrightarrowa connective, evaluated row by row

Worked example

Show that p→qp \to q and ∼p∨q\sim p \vee q are logically equivalent.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q140Moderate

Example 1 · Mathematical Logic · Logical Equivalence and Algebra of Statements

If statement I: If the work is not finished on time, the contractor is in trouble. Statement II: Either the work is finished on time or the contractor is in trouble. Then

Checking one row and declaring equivalence

Agreeing in a row proves nothing; equivalence needs EVERY row. The asymmetry is worth holding onto: one row can disprove equivalence but never establish it.

Concept 2 of 6: The Algebra of Statements

These laws are the ordinary algebra of numbers with AND behaving like multiplication and OR like addition — except that distribution works both ways round here, which ordinary arithmetic does not allow.

Definition

The full working list, gathered in one place. T denotes a tautology and F a contradiction:

  • Three of these you have already met as concepts in their own right — De Morgan and the negation of a conditional in Negation of Statements, and the conditional law itself back in the first block. They are repeated here because simplification uses them as ALGEBRA, one rewrite among ten, rather than as rules for negating.
  • The two that do the most work are complement and distributive.
  • The conditional law is what lets you remove an arrow so the others can apply.
  • Absorption is the one students most often fail to spot, and it collapses a whole bracket in one move.
LawWith ANDWith OR
Commutativep∧q≡q∧pp \wedge q \equiv q \wedge pp∨q≡q∨pp \vee q \equiv q \vee p
Associative(p∧q)∧r≡p∧(q∧r)(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)(p∨q)∨r≡p∨(q∨r)(p \vee q) \vee r \equiv p \vee (q \vee r)
Distributivep∧(q∨r)≡(p∧q)∨(p∧r)p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)p∨(q∧r)≡(p∨q)∧(p∨r)p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)
Both directions are legal here, unlike ordinary arithmetic where only one distribution holds.
Identityp∧T≡pp \wedge T \equiv pp∨F≡pp \vee F \equiv p
Dominationp∧F≡Fp \wedge F \equiv Fp∨T≡Tp \vee T \equiv T
Complementp∧∼p≡Fp \wedge \sim p \equiv Fp∨∼p≡Tp \vee \sim p \equiv T
The engine of most simplifications: spot a letter meeting its own negation and a whole branch collapses to F or T.
Idempotentp∧p≡pp \wedge p \equiv pp∨p≡pp \vee p \equiv p
Absorptionp∧(p∨q)≡pp \wedge (p \vee q) \equiv pp∨(p∧q)≡pp \vee (p \wedge q) \equiv p
The whole bracket vanishes. Worth memorising by shape: a letter outside meeting itself inside swallows the rest.
De Morgan∼(p∧q)≡  ∼p∨∼q\sim(p \wedge q) \equiv \;\sim p \vee \sim q∼(p∨q)≡  ∼p∧∼q\sim(p \vee q) \equiv \;\sim p \wedge \sim q
Conditional∼(p→q)≡p∧∼q\sim(p \to q) \equiv p \wedge \sim qp→q≡  ∼p∨qp \to q \equiv \;\sim p \vee q
Always apply this first. The other laws cannot see through an arrow.
Ten laws cover every simplification the paper sets. Complement, distributive and absorption account for most of the work.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q102Moderate

Example 2 · Mathematical Logic · Logical Equivalence and Algebra of Statements

The proposition (∼p)∨(p∧∼q)(\sim p) \vee (p \wedge \sim q) is equivalent to

Distributing only one way

In logic, OR distributes over AND as well: p∨(q∧r)≡(p∨q)∧(p∨r)p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r). Students trained on ordinary algebra expect only p∧(q∨r)p \wedge (q \vee r) to expand and miss the other half.
Drill 1 more on the algebra of statements

Concept 3 of 6: Simplifying a Statement Pattern

There is a reliable order: clear the arrows, push the negations in, then look for a letter meeting its own negation. Almost every 'is equivalent to' question on this paper collapses within three or four lines if you work in that order.

Definition

The standard simplification order:

  • Clear arrows with the conditional law, so only ∼,∧,∨\sim, \wedge, \vee remain.
  • Push negations inwards with De Morgan, cancelling double negations as they appear.
  • Distribute or factor to bring like letters together.
  • Collapse with complement, absorption, identity and domination.

Stop as soon as the expression matches an option — you do not have to reach the simplest possible form.

Worked example

Simplify [p∧(q∨r)]∨[∼r∧∼q∧p][p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p].
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q140Hard

Example 3 · Mathematical Logic · Logical Equivalence and Algebra of Statements

The logical statement (∼(∼p∨q)∨(p∧r))∧(∼q∧r)(\sim(\sim p\vee q)\vee(p\wedge r))\wedge(\sim q\wedge r) is equivalent to

Simplifying around an arrow instead of clearing it

De Morgan, distribution and absorption are stated for ∧\wedge and ∨\vee only. Leaving a →\to in place and trying to distribute across it produces confident nonsense. Clear every arrow first.
Drill 6 more on simplifying a statement pattern

Concept 4 of 6: Classifying a Pattern Without Building Its Table

The three classifications were defined in the first block as readings of the last column. The point here is that you rarely need the column: simplify with the laws and watch what falls out — a bare TT, a bare FF, or something still carrying a letter.

Definition

The classification is unchanged; what changes is how you reach it:

  • A tautology simplifies to TT — no letters survive.
  • A contradiction simplifies to FF — likewise.
  • A contingency simplifies to something that still contains a letter, which is the tell: if a letter survives, the value must depend on it, so the column cannot be constant.

That last line is the whole shortcut. You do not need to find the rows that differ — you only need to see that a letter is left.

Classification by last column

tautology⇒all Tcontradiction⇒all Fcontingency⇒mixed\text{tautology} \Rightarrow \text{all } T \qquad \text{contradiction} \Rightarrow \text{all } F \qquad \text{contingency} \Rightarrow \text{mixed}
  • p∨∼pp \vee \sim pthe standard tautology
  • p∧∼pp \wedge \sim pthe standard contradiction

Worked example

Classify (p∧∼q)∧(∼p∧q)(p \wedge \sim q) \wedge (\sim p \wedge q) as a tautology, contradiction or contingency.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q107Moderate

Example 4 · Mathematical Logic · Logical Equivalence and Algebra of Statements

If pp and qq are statements, then which of the following is a contingency.

Calling a contingency a tautology after checking two rows

A pattern that comes out T in the rows you happened to try may still be F elsewhere. Either simplify all the way to a bare TT, or check every row — a partial table cannot establish a tautology.
Drill 3 more on classifying a pattern without building its table

Concept 5 of 6: Finding the Statement That Makes a Pattern a Tautology

A recurring stem says 'this pattern is a tautology — what must r be?'. Turn it round: a conditional is a tautology exactly when its consequent is true in every row where the antecedent is. So simplify the antecedent first and read off what it forces.

Definition

For X→rX \to r to be a tautology, rr must be true in every row where XX is true.

  • Simplify XX first — it usually collapses to something small.
  • Read off the rows where the simplified XX holds.
  • The answer is the option that is true in all of those rows; it need not be true everywhere else.

A useful simplification: (p→q)∧∼q≡  ∼p∧∼q(p \to q) \wedge \sim q \equiv \;\sim p \wedge \sim q, the modus tollens pattern.

Tautology condition and modus tollens

(X→r) is a tautology  ⟺  X=T⇒r=T(p→q)∧∼q≡  ∼p∧∼q(X \to r) \text{ is a tautology} \iff X = T \Rightarrow r = T \qquad (p \to q) \wedge \sim q \equiv\; \sim p \wedge \sim q
  • XXthe antecedent — simplify this before anything else
  • rrthe statement being solved for

Worked example

The statement pattern [(p→q)∧∼q]→r[(p \to q) \wedge \sim q] \to r is a tautology. What must rr be equivalent to?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q109Moderate

Example 5 · Mathematical Logic · Logical Equivalence and Algebra of Statements

The statement pattern [(p→q)∧−q]→r\lbrack(p\rightarrow q) \land -q\rbrack \rightarrow r is a tautology when r is equivalent to

Looking for an r that is true everywhere

rr only has to hold in the rows where the ANTECEDENT is true — it may be false elsewhere without harming the tautology. Hunting for a universally true rr rules out the correct option, which is usually a plain ∼p\sim p or ∼q\sim q.
Drill 1 more on finding the statement that makes a pattern a tautology

Concept 6 of 6: The Dual of a Statement Pattern

The dual is a mechanical mirror: swap every AND with OR and every T with F, and leave the letters completely alone. It looks like a negation at a glance, and that resemblance is exactly what the paper tests.

Definition

The dual of a statement pattern is obtained by interchanging:

  • ∧\wedge with ∨\vee, and
  • TT with FF.

The statement letters and every ∼\sim stay exactly as they are. A conditional must be written in ∼,∧,∨\sim, \wedge, \vee form before a dual can be taken. The dual of the dual returns the original, and the laws in the table above come in dual pairs — which is why they are listed in two columns.

Dual versus negation

dual(p∧∼q)=p∨∼q∼(p∧∼q)≡  ∼p∨q\text{dual}(p \wedge \sim q) = p \vee \sim q \qquad \sim(p \wedge \sim q) \equiv\; \sim p \vee q
  • dualswap the connectives only — letters untouched
  • ∼\simnegation — swaps connectives AND negates every letter

Worked example

Write the dual of [∼q∧(p∨∼q)∧∼r]∨p[\sim q \wedge (p \vee \sim q) \wedge \sim r] \vee p.
Practise this conceptself-check · 4 quick reps

Negating the letters when asked for a dual

The dual of p∧∼qp \wedge \sim q is p∨∼qp \vee \sim q, NOT ∼p∨q\sim p \vee q. The second is the negation. Duals touch connectives and constants only, and the two operations produce different statements that both appear in the option list.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

  • Why Equivalence Licenses Substitution

    Equivalence and its test

    A≡B  ⟺  (A↔B) is a tautology  ⟺  A and B share every rowA \equiv B \iff (A \leftrightarrow B) \text{ is a tautology} \iff A \text{ and } B \text{ share every row}
  • Classifying a Pattern Without Building Its Table

    Classification by last column

    tautology⇒all Tcontradiction⇒all Fcontingency⇒mixed\text{tautology} \Rightarrow \text{all } T \qquad \text{contradiction} \Rightarrow \text{all } F \qquad \text{contingency} \Rightarrow \text{mixed}
  • Finding the Statement That Makes a Pattern a Tautology

    Tautology condition and modus tollens

    (X→r) is a tautology  ⟺  X=T⇒r=T(p→q)∧∼q≡  ∼p∧∼q(X \to r) \text{ is a tautology} \iff X = T \Rightarrow r = T \qquad (p \to q) \wedge \sim q \equiv\; \sim p \wedge \sim q
  • The Dual of a Statement Pattern

    Dual versus negation

    dual(p∧∼q)=p∨∼q∼(p∧∼q)≡  ∼p∨q\text{dual}(p \wedge \sim q) = p \vee \sim q \qquad \sim(p \wedge \sim q) \equiv\; \sim p \vee q

Reference tables (1)

The Algebra of Statements10 rows
LawWith ANDWith OR
Commutativep∧q≡q∧pp \wedge q \equiv q \wedge pp∨q≡q∨pp \vee q \equiv q \vee p
Associative(p∧q)∧r≡p∧(q∧r)(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)(p∨q)∨r≡p∨(q∨r)(p \vee q) \vee r \equiv p \vee (q \vee r)
Distributivep∧(q∨r)≡(p∧q)∨(p∧r)p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)p∨(q∧r)≡(p∨q)∧(p∨r)p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)
Both directions are legal here, unlike ordinary arithmetic where only one distribution holds.
Identityp∧T≡pp \wedge T \equiv pp∨F≡pp \vee F \equiv p
Dominationp∧F≡Fp \wedge F \equiv Fp∨T≡Tp \vee T \equiv T
Complementp∧∼p≡Fp \wedge \sim p \equiv Fp∨∼p≡Tp \vee \sim p \equiv T
The engine of most simplifications: spot a letter meeting its own negation and a whole branch collapses to F or T.
Idempotentp∧p≡pp \wedge p \equiv pp∨p≡pp \vee p \equiv p
Absorptionp∧(p∨q)≡pp \wedge (p \vee q) \equiv pp∨(p∧q)≡pp \vee (p \wedge q) \equiv p
The whole bracket vanishes. Worth memorising by shape: a letter outside meeting itself inside swallows the rest.
De Morgan∼(p∧q)≡  ∼p∨∼q\sim(p \wedge q) \equiv \;\sim p \vee \sim q∼(p∨q)≡  ∼p∧∼q\sim(p \vee q) \equiv \;\sim p \wedge \sim q
Conditional∼(p→q)≡p∧∼q\sim(p \to q) \equiv p \wedge \sim qp→q≡  ∼p∨qp \to q \equiv \;\sim p \vee q
Always apply this first. The other laws cannot see through an arrow.
Ten laws cover every simplification the paper sets. Complement, distributive and absorption account for most of the work.

Watch out for (6)

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