0
0

Coded / Symbolic Syllogism

Introduction

Coded / Symbolic Syllogisms replace natural-language quantifiers with compact symbols (for example, +, -, ×) so you can focus on structure and inference rules quickly. These questions are common in modern aptitude tests because they test the same syllogistic reasoning but in a speed-friendly shorthand.

Learning to decode symbols into logical relations, apply syllogistic rules, and then re-encode the conclusion is the core skill for this pattern.

Pattern: Coded / Symbolic Syllogism

Pattern

The key idea: Translate each coded statement into its logical form (All, No, Some), perform ordinary syllogistic inference, then map the valid conclusion(s) back to symbols.

Common symbol mapping (use the test’s legend if provided - these are typical):

  • A + B → All A are B (universal affirmative)
  • A - B → No A is B (universal negative)
  • A × B → Some A are B (particular affirmative)
  • A ÷ B or A ~ B - sometimes used for Some A are not B (particular negative); check the question legend.
Rules to apply:
  • Always decode symbols to natural language first - work with All / No / Some logic.
  • Apply standard syllogistic checks (distribution of middle, negative-negative prohibition, figure/mood awareness).
  • Only after determining which natural-language conclusions hold, re-encode into the test’s symbolic form for the answer.
  • If the legend is ambiguous, treat symbols consistently across the problem and state your decoded mapping mentally before solving.

Step-by-Step Example

Question

Legend: + = All, - = No, × = Some.
Statements:
1️⃣ P + Q
2️⃣ Q × R

Which coded conclusion is valid?
Options:
A. P × R    B. P - R    C. R + P    D. P + R

Solution

  1. Step 1: Decode the symbols

    P + Q ⇒ All P are Q.
    Q × R ⇒ Some Q are R.

  2. Step 2: Draw natural-language inference

    From All P are Q and Some Q are R, we can validly infer Some P are R (the particular part of Q that overlaps R might include some P). This is a standard A-I → I pattern (Universal + Particular → Particular).

  3. Step 3: Re-encode the valid conclusion

    Some P are R → symbolically P × R.

  4. Final Answer:

    P × R → Option A

  5. Quick Check:

    If all P sit inside Q and part of Q overlaps R, some of those P may lie in that overlapping part → Some P are R holds. ✅

Quick Variations

1. A-A chains: A + B and B + C → A + C (All → All).

2. Negative involvement: A negative symbol (-) anywhere often forces particular-negative or blocks transitive universals. Decode carefully.

3. Existential limits: If the coded system includes a symbol for “Some not” (e.g., ÷), decode it as particular negative and obey existential rules.

4. Figure awareness: when two universals are present, check if middle term distribution allows universal transitivity.

Trick to Always Use

  • Step 1 → Always expand every coded premise into plain English before manipulating.
  • Step 2 → Mark the middle term and check if it’s distributed in at least one premise (to avoid undistributed middle errors).
  • Step 3 → Translate the final natural-language conclusion back into the same symbols used by the question (consistency matters).

Summary

Summary

  • Decode symbols into All / No / Some before reasoning.
  • Ensure the middle term is distributed as required; avoid chaining through an undistributed middle.
  • Re-encode only those conclusions that survive standard syllogistic checks.
  • When in doubt, draw a quick Venn sketch for the decoded premises - then translate the visual back into symbols.

Example to remember:
A + B; B × C ⇒ A × C - All A are B and Some B are C gives Some A are C (symbolically: A + B; B × C ⇒ A × C). ✅

Practice

(1/5)
1. Legend: '+' = All, '-' = No, '×' = Some.<br>Statements: 1️⃣ M + N, 2️⃣ N × O. What follows?
easy
A. <code>M × O</code>
B. <code>M - O</code>
C. <code>O + M</code>
D. <code>M + O</code>

Solution

  1. Step 1: Decode

    M + N ⇒ All M are N; N × O ⇒ Some N are O.

  2. Step 2: Infer

    All M are N and Some N are O ⇒ Some M are O (particular positive).

  3. Step 3: Encode back

    ‘Some M are O’ ⇒ M × O.

  4. Final Answer:

    M × O → Option A

  5. Quick Check:

    Universal + Particular = valid Some. ✅
Hint: A + B and B × C ⇒ A × C
Common Mistakes: Assuming universal result instead of particular.
2. Legend: '+' = All, '-' = No, '×' = Some.<br>Statements: 1️⃣ P - Q, 2️⃣ Q + R. Which coded conclusion follows?
easy
A. <code>P × R</code>
B. <code>P - R</code>
C. <code>R - P</code>
D. None of the above

Solution

  1. Step 1: Decode the premises

    P - Q ⇒ No P is Q.
    Q + R ⇒ All Q are R.

  2. Step 2: Analyze possible relations

    The premises tell us P is disjoint from the Q-region, and Q is fully inside R. However, R may have regions outside Q. P could overlap those parts of R that are outside Q, or it might not - the premises do not fix that. Therefore none of the definite relations A-C can be concluded.

  3. Step 3: Conclusion

    No definite coded conclusion among the given options follows from the premises.

  4. Final Answer:

    None of the above. → Option D

  5. Quick Check:

    No + All does not force a No or an All between P and R because R can include non-Q elements where P may or may not lie. ✅
Hint: No A is B and All B are C ⇒ you cannot assume relation between A and C unless R is limited to B.
Common Mistakes: Treating 'All Q are R' as 'All R are Q' (reverse) and inferring P-R exclusion incorrectly.
3. Legend: '+' = All, '-' = No, '×' = Some.<br>Statements: 1️⃣ A + B, 2️⃣ B - C. Which of the following follows?
easy
A. <code>A × C</code>
B. <code>A - C</code>
C. <code>C + A</code>
D. <code>A + C</code>

Solution

  1. Step 1: Decode

    A + B ⇒ All A are B; B - C ⇒ No B is C.

  2. Step 2: Infer

    If All A are B and No B is C ⇒ No A is C (A-E → E form).

  3. Step 3: Encode

    No A is C ⇒ A - C.

  4. Final Answer:

    A - C → Option B

  5. Quick Check:

    All + No ⇒ No. ✅
Hint: All + No ⇒ No (A-E → E).
Common Mistakes: Assuming possible overlap despite explicit exclusion.
4. Legend: '+' = All, '-' = No, '×' = Some, '÷' = Some not.<br>Statements: 1️⃣ L × M, 2️⃣ M + N. Which conclusion follows?
medium
A. <code>L × N</code>
B. <code>L + N</code>
C. <code>L - N</code>
D. <code>L ÷ N</code>

Solution

  1. Step 1: Decode

    L × M ⇒ Some L are M; M + N ⇒ All M are N.

  2. Step 2: Infer

    Some L are M and All M are N ⇒ Some L are N (I-A → I form).

  3. Step 3: Encode back

    ‘Some L are N’ ⇒ L × N.

  4. Final Answer:

    L × N → Option A

  5. Quick Check:

    Some + All ⇒ Some valid. ✅
Hint: Some + All ⇒ Some.
Common Mistakes: Treating partial overlap as universal.
5. Legend: '+' = All, '-' = No, '×' = Some.<br>Statements: 1️⃣ S + T, 2️⃣ T - U. What coded conclusion follows?
medium
A. <code>S - U</code>
B. <code>S × U</code>
C. <code>U - S</code>
D. <code>S + U</code>

Solution

  1. Step 1: Decode

    S + T ⇒ All S are T; T - U ⇒ No T is U.

  2. Step 2: Infer

    All S are T and No T is U ⇒ No S is U.

  3. Step 3: Encode back

    ‘No S is U’ ⇒ S - U.

  4. Final Answer:

    S - U → Option A

  5. Quick Check:

    All + No ⇒ No. ✅
Hint: All + No ⇒ No conclusion of inclusion, always exclusion.
Common Mistakes: Confusing ‘No’ with ‘Some not’.

Mock Test

Ready for a challenge?

Take a 10-minute AI-powered test with 10 questions (Easy-Medium-Hard mix) and get instant SWOT analysis of your performance!

10 Questions
5 Minutes