Models part 4

1. MODELS
PART 4

2. HW ANSWERS

3. PART F

4. • ∃xPx → Qc, ∃x (Px → Qc)
UD = {Marry, Austin}
Ext (P) = {
Ref (Q) = {
Ref (c) = Marry
When is ∃x (Px → Qc) false?
1st case
When x is in P and c is not in Q.
Would it be false if nothing is in P? No, it actually will be true. If nothing is in P, it will be true,
and if something is not in P, it will also be true.
∃x (Px → Qc) means (Pm →Qc) v (Pa →Qc)
If any object of UD is not in P, the whole formula will be TRUE!
(Pm →Qc) v (Pa →Qc)
0 1 1/0 1 1 1/0

5. • So, another case when this formula is false is when everything (everything from UD) is in P and c is not in
Q
(Pm →Qc) v (Pa →Qc)
1 0 0 1 0 0
_________________
∃xPx → Qc, ∃x (Px → Qc)
0 1
UD = {Marry, Austin}
Ext (P) = {Marry}
Ref (Q) = {0}
Ref (c) = Marry

6. 5. ∀x (Px → ¬ Qx), ∃x (Px & ¬ Qx)
1 0
∃x (Px & ¬ Qx) = (Pvika& ¬ Qvika) v (Pivan & ¬Qivan)
To be false both disjuncts has to be false
UD: {Vika, Ivan}
Ext. (P) = {0}
Ext (Q) = {Vika}

7. PART E #4
4. ∀z Jz ↔ ∃yJy
This formula is false in this model:
∀z Jz ↔ ∃y Jy
0 1
UD= {Vika, Ivan}
Ext (J) ={Vika}
This formula is true in this model:
∀z Jz ↔ ∃y Jy
1 1
UD= {Vika, Ivan}
Ext (J) ={Vika, Ivan}

8. CONSISTENCY
• A SET OF SENTENCES IS CONSISTENT
WHEN THERE IS A MODEL WHERE ALL OF
THEM ARE TRUE

9. PRACTICE

10. PART G #1
• {Ma, ¬ Na, Pa, ¬ Qa}
1 1 1 1
UD = {Angie}
Ext (M) = {Angie}
Ext (N) = {0}
Ext (P) ={Angie}
Ext (Q) = {0}
Ref (a) = Angie

11. VALIDITY IN QL
• A SET OF SENTENCES IS VALID IFF THERE IS NO
MODEL WHERE THE PREMISES ARE TRUE AND THE
CONCLUSION IS FALSE. OTHERWISE, THE SET OF
SENTENCES IS INVALID

12. TO PROVE VALIDITY…
WE NEED TO SHOW THAT THERE IS A MODEL
WHERE PREMISES ARE TRUE AND THE
CONCLUSION IS FALSE

13. PART H
1. ∀x (Ax → Bx) ∴ ∃xBx
1 0
UD: {Sasha, Masha}
Ext (A) = {0}
Ext (B) = {0}