M4 2.5 Truth Table worksheet - Free Printable

We are given the logical statement:
¬(p → r) ∨ q

We need to complete the truth table by evaluating this expression for all combinations of truth values for p, q, and r.

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Step 1: Understand the components

We have three atomic propositions: p, q, r — each can be True (T) or False (F).
There are 2³ = 8 possible combinations.

The expression is:
¬(p → r) ∨ q

This breaks down into:
- First, compute (p → r) — the implication.
- Then, negate it: ¬(p → r)
- Finally, OR that with q: ¬(p → r) ∨ q

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Step 2: Recall truth tables for basic operations

#### Implication (→):
| p | r | p → r |
|---|---|--------|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |

> Note: The only time p → r is false is when p is true and r is false.

#### Negation (¬):
| X | ¬X |
|---|----|
| T | F |
| F | T |

#### Disjunction (∨):
| A | B | A ∨ B |
|---|---|--------|
| T | T | T |
| T | F | T |
| F | T | T |
| F | F | F |

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Step 3: Compute column by column for all 8 rows

We’ll go row by row.

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#### Row 1: p=T, q=T, r=T
- (p → r) = T → T = T
- ¬(p → r) = ¬T = F
- ¬(p → r) ∨ q = F ∨ T = T

#### Row 2: p=T, q=T, r=F
- (p → r) = T → F = F
- ¬(p → r) = ¬F = T
- ¬(p → r) ∨ q = T ∨ T = T

#### Row 3: p=T, q=F, r=T
- (p → r) = T → T = T
- ¬(p → r) = ¬T = F
- ¬(p → r) ∨ q = F ∨ F = F

#### Row 4: p=T, q=F, r=F
- (p → r) = T → F = F
- ¬(p → r) = ¬F = T
- ¬(p → r) ∨ q = T ∨ F = T

#### Row 5: p=F, q=T, r=T
- (p → r) = F → T = T
- ¬(p → r) = ¬T = F
- ¬(p → r) ∨ q = F ∨ T = T

#### Row 6: p=F, q=T, r=F
- (p → r) = F → F = T
- ¬(p → r) = ¬T = F
- ¬(p → r) ∨ q = F ∨ T = T

#### Row 7: p=F, q=F, r=T
- (p → r) = F → T = T
- ¬(p → r) = ¬T = F
- ¬(p → r) ∨ q = F ∨ F = F

#### Row 8: p=F, q=F, r=F
- (p → r) = F → F = T
- ¬(p → r) = ¬T = F
- ¬(p → r) ∨ q = F ∨ F = F

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Final Completed Truth Table:

| p | q | r | (p → r) | ¬(p → r) | ¬(p → r) ∨ q |
|---|---|---|---------|----------|--------------|
| T | T | T | T | F | T |
| T | T | F | F | T | T |
| T | F | T | T | F | F |
| T | F | F | F | T | T |
| F | T | T | T | F | T |
| F | T | F | T | F | T |
| F | F | T | T | F | F |
| F | F | F | T | F | F |

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✔ This completes the truth table for the statement ¬(p → r) ∨ q.

Let me know if you’d like to see a simplified version or analyze when this statement is true/false!