Math Teacher Mambo: Truth Tables - Free Printable

Let's solve each of these truth table problems step by step, using basic logical rules:

Logical Operators Review:

- AND (∧): True only if both operands are true.
- OR (∨): True if at least one operand is true.
- NOT (~): Negates the truth value.
- IMPLIES (→): False only when the antecedent is T and consequent is F.
- IFF (↔): True when both sides have the same truth value.
- XOR (⊕): True when exactly one side is true.

We’ll go through each problem and fill in the missing column based on the given values.

---

6.

Given:
| p | w | ~(p ∧ w) |
|---|---|----------|
| F | T | |
| T | T | |
| T | F | |

First compute `p ∧ w`, then negate it.

1. F ∧ T = F → ~F = T
2. T ∧ T = T → ~T = F
3. T ∧ F = F → ~F = T

✔ Answer: T, F, T

---

7.

Given:
| p | (w ∨ v) | p → (w ∨ v) |
|---|--------|-------------|
| F | T | |
| T | T | |
| T | F | |
| F | T | |

We need to evaluate `p → (w ∨ v)`.

Recall: `A → B` is false only when A=T and B=F.

1. F → T = T
2. T → T = T
3. T → F = F
4. F → T = T

✔ Answer: T, T, F, T

---

8.

Given:
| (y ∧ t) | r | (y ∧ t) ∨ r |
|--------|---|-------------|
| T | F | |
| F | T | |
| F | F | |
| T | T | |

Compute `(y ∧ t) ∨ r`

1. T ∨ F = T
2. F ∨ T = T
3. F ∨ F = F
4. T ∨ T = T

✔ Answer: T, T, F, T

---

9.

Given:
| (k ↔ p) | g | g ∧ (k ↔ p) |
|--------|---|-------------|
| F | T | |
| T | T | |
| T | F | |

Evaluate `g ∧ (k ↔ p)`

1. T ∧ F = F
2. T ∧ T = T
3. F ∧ T = F

✔ Answer: F, T, F

---

10.

Given:
| ~w ∧ r | t | (~w ∧ r) ↔ t |
|--------|---|--------------|
| T | F | |
| F | T | |
| T | F | |
| T | T | |

We need to evaluate `(~w ∧ r) ↔ t`

This is true when both sides are equal.

1. T ↔ F = F
2. F ↔ T = F
3. T ↔ F = F
4. T ↔ T = T

✔ Answer: F, F, F, T

---

11.

Given:
| ~b | (k ∨ p) | (k ∨ p) ∧ ~b |
|----|---------|---------------|
| T | F | |
| F | F | |
| F | T | |
| T | F | |

Evaluate `(k ∨ p) ∧ ~b`

1. F ∧ T = F
2. F ∧ F = F
3. T ∧ F = F
4. F ∧ T = F

✔ Answer: F, F, F, F

---

12.

Given:
| (n ↔ j) | e | (n ↔ j) ∨ e |
|--------|---|-------------|
| F | T | |
| F | F | |
| T | T | |
| T | F | |

Evaluate `(n ↔ j) ∨ e`

1. F ∨ T = T
2. F ∨ F = F
3. T ∨ T = T
4. T ∨ F = T

✔ Answer: T, F, T, T

---

13.

Given:
| u | (~h ∧ c) | (~h ∧ c) → u |
|---|----------|----------------|
| T | T | |
| F | F | |
| T | F | |
| F | T | |

Evaluate `(~h ∧ c) → u`

1. T → T = T
2. F → F = T (since implication is only false when antecedent T and consequent F)
3. F → T = T
4. T → F = F

✔ Answer: T, T, T, F

---

14.

Given:
| (w ∨ ~r) | ~(w ∨ ~r) |
|----------|-----------|
| T | |
| F | |
| T | |

Negate the first column:

1. ~T = F
2. ~F = T
3. ~T = F

✔ Answer: F, T, F

---

15.

Given:
| f → r | ~t | (f → r) ↔ ~t |
|-------|----|--------------|
| T | F | |
| F | T | |
| F | F | |
| T | T | |

Evaluate `(f → r) ↔ ~t`

So compare the two columns:

1. T ↔ F = F
2. F ↔ T = F
3. F ↔ F = T
4. T ↔ T = T

✔ Answer: F, F, T, T

---

16.

Given:
| (j ∧ u) | ~y | ~y → (j ∧ u) | (j ∧ u) ∨ ~y |
|--------|----|--------------|----------------|
| T | F | | |
| F | T | | |
| T | T | | |

We need to fill two columns:
- `~y → (j ∧ u)`
- `(j ∧ u) ∨ ~y`

#### First: `~y → (j ∧ u)`

1. ~y = F, (j ∧ u) = T → F → T = T
2. ~y = T, (j ∧ u) = F → T → F = F
3. ~y = T, (j ∧ u) = T → T → T = T

So: T, F, T

#### Second: `(j ∧ u) ∨ ~y`

1. T ∨ F = T
2. F ∨ T = T
3. T ∨ T = T

So: T, T, T

✔ Final answers:
- `~y → (j ∧ u)` → T, F, T
- `(j ∧ u) ∨ ~y` → T, T, T

---

✔ Final Answer Bank (Matching with the provided answer bank):

Let’s compile all results in order:

1. 6: T, F, T
2. 7: T, T, F, T
3. 8: T, T, F, T
4. 9: F, T, F
5. 10: F, F, F, T
6. 11: F, F, F, F
7. 12: T, F, T, T
8. 13: T, T, T, F
9. 14: F, T, F
10. 15: F, F, T, T
11. 16:
- `~y → (j ∧ u)` → T, F, T
- `(j ∧ u) ∨ ~y` → T, T, T

Now match to answer bank (which shows a grid of possible answers):

The answer bank has multiple options, but we can now fill in the blanks as follows:

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✔ Final Answers for Each Problem:

| # | Answer |
|---|--------|
| 6 | T, F, T |
| 7 | T, T, F, T |
| 8 | T, T, F, T |
| 9 | F, T, F |
|10 | F, F, F, T |
|11 | F, F, F, F |
|12 | T, F, T, T |
|13 | T, T, T, F |
|14 | F, T, F |
|15 | F, F, T, T |
|16 | T, F, T (for implication), T, T, T (for disjunction) |

Since the answer bank appears to be a collection of truth value sequences, you can now match each row accordingly.

For example, the first column of the answer bank is:
```
T
F
T
```
Which matches Problem 6.

Similarly, second column:
```
T
T
F
T
```
Matches Problem 7, etc.

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✔ Summary:

All truth tables are solved using basic logic rules. The key is understanding:
- Implication (`→`) is false only when T → F
- Biconditional (`↔`) is true when both sides match
- AND requires both true
- OR is true if at least one is true
- NOT flips the value

You can now fill in the truth tables correctly!

Let me know if you want the full filled-in tables!