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MODELS
PART 3
HW ANSWERS
PART B #8-10
PART B #5-6
1. ∃xTxh
Step 1. Let’s prove prove that this formula is not a tautology.
We need a model where this senetnece is false.
This formula is false in this model:
Step 2. Let’s set the UD:
UD = {Abby, Billy}
Step 3. Now let’s set the referent for the singular terms:
In the formula we have ‘h’ as a singular term.
What will be there referent for ‘h’?
We can set up Abby as the referent for h.
Referent (h) = Abby
Step 4. Now let’s set the extension for T in the way that would make this formula false:
Extension (T) = {0}
∃xTxh
Step 5. Let’s show now that this formula is not a contradiction.
It’s not a contradiction if there is a model where this formula is true.
Let’s construct a model where this formula is true,
This formula is true in this model:
UD= {Abby, Billy}
Extension (T)= {<Billy, Abby>}
Referent (h)= Abby
We have shown that this formula is not a tautology – because have shown that there is a
model where it’s false.
And we have shown that this formula is not a contradiction by providing the model where it’s
true.
It’s not a contradiction and it’s not a tautology – therefore, it’s contingent.
LOGICAL EQUIVALENCY
• A SET OF SENTENCES IS LOGICALLY EQUIVALENT WHEN A |= B
and B |= A
A |= B means that there is no model where A is true and B is false
B |= A means that there is no model where B is true and A is false.
A SET OF SENTENCES IS NOT LOGICALLY EQUIVALENT WHEN THERE IS A MODEL WHERE
EITHER A IS TRUE AND B IS FALSE, OR A MODEL WHERE B IS TRUE AND A IS FALSE
TO SHOW THAT SENTENCES ARE NOT
LOGICALLY EQUIVALENT ….
• To show that a set of sentences is not logically equivalent we need to
show a model where one sentence is true and another is false.
PRACTICE: PART F
• 1. Ja, Ka
To show that they are not logically equivalent we need to construct a model where one of the sentences is
true and another is false. For example, Ja is true and Ka is false.
Step 1. Ja, Ka
1 0
UD: {Abby, Billy}
Extension (J):
Extension (K):
Referent (a): Abby
UD: {Abby, Billy}
Extension (J): {Abby}
Extension (K): {0}
Referent (a): Abby
• ∃xJx, Jm
1 0
UD: {Marry, Austin}
Ext (J) = {Austin}
Ref (m) = Marry
• ∀xRxx, ∃xRxx
0 1
UD: {Marry, Austin}
Ext (R) = { <Marry, Marry>, <Marry, Austin>}
• ∃xPx → Qc, ∃x (Px → Qc)
0 1
UD = {Marry, Austin}
Ext (P) = {Austin}
Ext (Q) = {0}
Ref (c) = Marry
CONSISTENCY
• A SET OF SENTENCES IS CONSISTENT WHEN THERE IS A MODEL
WHERE ALL OF THEM ARE TRUE

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Models part 3

  • 5. 1. ∃xTxh Step 1. Let’s prove prove that this formula is not a tautology. We need a model where this senetnece is false. This formula is false in this model: Step 2. Let’s set the UD: UD = {Abby, Billy} Step 3. Now let’s set the referent for the singular terms: In the formula we have ‘h’ as a singular term. What will be there referent for ‘h’? We can set up Abby as the referent for h. Referent (h) = Abby Step 4. Now let’s set the extension for T in the way that would make this formula false: Extension (T) = {0}
  • 6. ∃xTxh Step 5. Let’s show now that this formula is not a contradiction. It’s not a contradiction if there is a model where this formula is true. Let’s construct a model where this formula is true, This formula is true in this model: UD= {Abby, Billy} Extension (T)= {<Billy, Abby>} Referent (h)= Abby We have shown that this formula is not a tautology – because have shown that there is a model where it’s false. And we have shown that this formula is not a contradiction by providing the model where it’s true. It’s not a contradiction and it’s not a tautology – therefore, it’s contingent.
  • 7. LOGICAL EQUIVALENCY • A SET OF SENTENCES IS LOGICALLY EQUIVALENT WHEN A |= B and B |= A A |= B means that there is no model where A is true and B is false B |= A means that there is no model where B is true and A is false. A SET OF SENTENCES IS NOT LOGICALLY EQUIVALENT WHEN THERE IS A MODEL WHERE EITHER A IS TRUE AND B IS FALSE, OR A MODEL WHERE B IS TRUE AND A IS FALSE
  • 8. TO SHOW THAT SENTENCES ARE NOT LOGICALLY EQUIVALENT …. • To show that a set of sentences is not logically equivalent we need to show a model where one sentence is true and another is false.
  • 9. PRACTICE: PART F • 1. Ja, Ka To show that they are not logically equivalent we need to construct a model where one of the sentences is true and another is false. For example, Ja is true and Ka is false. Step 1. Ja, Ka 1 0 UD: {Abby, Billy} Extension (J): Extension (K): Referent (a): Abby UD: {Abby, Billy} Extension (J): {Abby} Extension (K): {0} Referent (a): Abby
  • 10. • ∃xJx, Jm 1 0 UD: {Marry, Austin} Ext (J) = {Austin} Ref (m) = Marry • ∀xRxx, ∃xRxx 0 1 UD: {Marry, Austin} Ext (R) = { <Marry, Marry>, <Marry, Austin>}
  • 11. • ∃xPx → Qc, ∃x (Px → Qc) 0 1 UD = {Marry, Austin} Ext (P) = {Austin} Ext (Q) = {0} Ref (c) = Marry
  • 12. CONSISTENCY • A SET OF SENTENCES IS CONSISTENT WHEN THERE IS A MODEL WHERE ALL OF THEM ARE TRUE