---
title: "Tableau for (T∨S)→¬Q"
subject: "Discrete Structure"
chapter: 1
key: "semester_4/discrete_structure:tableau-t-or-s-implies-not-q"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/tableau-t-or-s-implies-not-q
site: BCT Engineering
---

# Tableau for (T∨S)→¬Q

Discrete Structure · Chapter 1: Logic, Induction and Reasoning

- Asked in: 2072 Asoj, 3 marks; 2068 Bhadra, 3 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/discrete_structure/tableau-t-or-s-implies-not-q
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-1

## Question

Draw the tableau for the formula (T∨S) → ¬Q where ¬ denotes the negation of variable, ∨ denotes the disjunction of variables and → is the symbol for implication.

## Short answer

The tableau for (T∨S)→¬Q branches: left produces ¬(T∨S) giving ¬T, ¬S (open); right gives ¬Q (open). No branch closes, so the formula is satisfiable but not valid. Models: {T=false, S=false} or {Q=false}.

## Answer

Formula: (T ∨ S) → ¬Q.

**Rule used.** A → B is a β-formula: it branches into ¬A | B. Then ¬(T ∨ S) is an α-formula: it gives ¬T and ¬S on the same branch.

#### Tableau
```text
          (T ∨ S) → ¬Q
            /      \
     ¬(T ∨ S)       ¬Q
         |           o
        ¬T
         |
        ¬S
         o
```

#### Reading the tableau
| Branch | Literals | Closed? |
|---|---|---|
| Left | ¬T, ¬S | No, open (o) |
| Right | ¬Q | No, open (o) |

Neither branch contains a complementary pair, so the tableau is **open** and the formula is **satisfiable**.

- Left branch gives the model T = false, S = false (Q can be anything).
- Right branch gives the model Q = false (T, S can be anything).

Since the tableau is open, the formula is not unsatisfiable. (It is not valid either, for example T = true, Q = true makes it false.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the first rule split (T ∨ S) → ¬Q?
   A. T ∨ S and ¬Q on the same branch
   B. ¬(T ∨ S) | ¬Q
   C. T | S
   D. ¬(T ∨ S) | Q
   Answer: B. A → B is a β-formula that branches into ¬A | B, here ¬(T ∨ S) | ¬Q. (Key point: Root split)

2. What happens to ¬(T ∨ S) on the left branch?
   A. It splits into ¬T | ¬S
   B. It gives ¬T and ¬S on the same branch
   C. It becomes T ∨ S
   D. It closes immediately
   Answer: B. ¬(A ∨ B) is an α-formula giving ¬A and ¬B on one branch. (Key point: Left branch)

3. What is the final status of the tableau?
   A. Closed, so the formula is unsatisfiable
   B. Open, so the formula is satisfiable
   C. Open, so the formula is valid
   D. Closed, so the formula is valid
   Answer: B. Neither branch has a complementary pair, so the tableau is open and the formula is satisfiable (not necessarily valid). (Key point: Conclusion)

## Close variants set in other exams

- [Tableau for a formula set](https://www.bctengineer.com/question.md?key=semester_4%2Fdiscrete_structure%3Atableau-formula-set) — Discrete Structure, chapter 1, asked 5×
- [Tableau for (p∧¬q)→s](https://www.bctengineer.com/question.md?key=semester_4%2Fdiscrete_structure%3Atableau-p-and-not-q-implies-s) — Discrete Structure, chapter 1, asked 5×
- [Tableau for ¬((p∧q)∨r)](https://www.bctengineer.com/question.md?key=semester_4%2Fdiscrete_structure%3Atableau-not-pq-or-r) — Discrete Structure, chapter 1, asked 5×