Logical “And”
Learning Target & Do Now
LT: By the end of today, I will know the definition of the truth functional connective “and”.
DN: Could we translate each of the following into our current logical language? Why or why not?
“A number x is divisible by 2 and 3.” “Tom will go to Berlin and Paris.”
“Steve is from Texas but not from Dallas.”
Conjunctions
P = Tom will go to Berlin Q = Tom will go to Paris P and Q
R = x is evenly divisible by 2 S = x is evenly divisible by 3 R and S
T = Steve is from Texas U = Steve is from Dallas T and ~U
Conjunctions, cont.
T = Steve is from Texas U = Steve is from Dallas T and ~U
Symbolic logic syntax: T ^ ~U
P ^ Q
P | Q | P ^ Q |
T | T | |
T | F | |
F | T | |
F | F | |
Alternative Phrasings
Some alternative phrasings are explicit:
Others, implicit. These usually state multiple properties/conditions:
Can we turn each of these into logical conjunctions?
English Traps with “And”
Conjunction Inference Rules
Practice
Tom and Steve will go to London. If Steve goes to London, then he will ride the Eye. Tom will ride the Eye too, provided he goes to London. So, both Steve and Tom will ride the Eye.
Make a translation key. What is our argument? Can we prove it is valid?
Solution
Tom will go to London = P Tom will ride the Eye = Q�Steve will go to London = R Steve will ride the Eye = S
1. P ^ R Premise 4. P Simplification�2. R → S Premise 5. R Simplification�3. P → Q Premise 6. S Modus Ponens, 2, 5� _________ 7. Q Modus Ponens, 3, 4�S ^ Q | 8. S ^ Q Adjunction, 6, 7
You Do
Prove the following:
(Q → ~S)�(P → (Q ^ R))�T → ~R�P�_____________
(~S ^ ~T)
Solution
1. (Q → ~S) Premise�2. (P → (Q ^ R)) Premise�3. T → ~R Premise�4. P Premise�____________________________________
5. Q ^ R Modus Ponens, 2, 4�6. Q Simplification, 5�7. ~S Modus Ponens, 1, 6�8. R Simplification, 5
9. ~~R Double Negation, 8�10. ~T Modus Tollens, 3, 9�11. ~S ^ ~T Adjunction, 7, 10
(This is only one of many possible solutions. You may have a different one!)
Homework
DeLancey 5.6 #1, 2
Today’s animal: pink fairy armadillo
Complex Sentences
A lesson, “and” then some!
Learning Target & Do Now
LT: By the end of today, I will have some experience working with sentences containing several connectives.
DN: Consider the following three sentences. How does the truth value(s) of each differ?
~(P → Q) (~P → Q) (~P → ~Q)
Parenthese Placement
How should we determine the truth of such sentences?
Truth Tables… But Bigger
Sample: ~(P → Q)
P | Q | P → Q | ~(P → Q) |
T | T | | |
T | F | | |
F | T | | |
F | F | | |
You Do
Create the truth tables for (~P → Q) and (~P → ~Q).
I recommend 4 and 5 columns, respectively
(~P → Q)
P | Q | ~P | (~P → Q) |
T | T | | |
T | F | | |
F | T | | |
F | F | | |
(~P → Q)
P | Q | ~P | (~P → Q) |
T | T | F | T |
T | F | F | T |
F | T | T | T |
F | F | T | F |
(~P → ~Q)
P | Q | ~P | ~Q | (~P → ~Q) |
T | T | | | |
T | F | | | |
F | T | | | |
F | F | | | |
(~P → ~Q)
P | Q | ~P | ~Q | (~P → ~Q) |
T | T | F | F | T |
T | F | F | T | T |
F | T | T | F | F |
F | F | T | T | T |
Validity of Complex Sentences
Homework
Today’s animal: Secretarybird
Conditional Derivations
Learning Target & Do Now
LT: By the end of class, I will see how conditional derivation can occur.
DN: Determine if the following argument is valid (simplification of a quote from Thomas Hobbes’ Leviathan):
If we want to be safe, then we should have a state that can protect us.�If we should have a state that can protect us, then we should give up some freedoms.
Therefore, if we want to be safe, we should give up some freedoms.
DN Semantics and Syntax
Semantically, the argument is valid (table from textbook).
Our conclusion is an implication - we have not yet seen how to prove these syntactically.
Conditional Derivation
The Relation of Proofs to Conditionals
At its core, a proof is a conditional statement
If all of the premises are true, then the conclusion should be true (roughly)
Therefore, a proof is a conditional!
Conditional Derivation
Conditional Derivation Example
You Do
Prove the following:
(P → Q)�(R → S)�____________
((P ^ R) → (Q ^ S))
Hint: assume the conjunction P^R
Solution ((P ^ R) → (Q ^ S))
1. P → Q premise 6. R Simplification, 2
2. R → S premise 7. S Modus Ponens, 2, 6
__________________________ 8. (Q ^ S) Adjunction, 5, 7
3. P ^ R Asmp. for conditional deriv 9. ((P ^ R) → (Q ^ S))
4. P Simplification, 1 Conditional Deriv, 3-8
5. Q Modus Ponens, 1, 4
Homework
DeLancey 6.5 #1
Today’s animal: Siphonophore
Tautologies and Theorems
Learning Target & Do Now
LT: By the end of today, I will know what tautologies and theorems are.
DN: Think-pair-share: what is conditional derivation?� With your partner, prove P → P.
A Tautology and a Theorem
Let us build the truth table for the following:
((P → Q) → (~Q → ~P))
P | Q | ~P | ~Q | (P → Q) | (~Q → ~P) | ((P → Q) → (~Q → ~P)) |
T | T | | | | | |
T | F | | | | | |
F | T | | | | | |
F | F | | | | | |
A Tautology and a Theorem
The expression is always true! This is a tautology.
((P → Q) → (~Q → ~P))
P | Q | ~P | ~Q | (P → Q) | (~Q → ~P) | ((P → Q) → (~Q → ~P)) |
T | T | F | F | T | T | T |
T | F | F | T | F | F | T |
F | T | T | F | T | T | T |
F | F | T | T | T | T | T |
A Tautology and a Theorem
Prove the following is valid:
((P → Q) → (~Q → ~P))
Remember: Assume the antecedent, prove the consequent
A Tautology and a Theorem
Prove the following is valid:
((P → Q) → (~Q → ~P))
1. P → Q Assumption for cond. deriv�2. ~Q Assumption for cond. deriv�3. ~P Modus Tollens, 1, 2�4. (~Q → ~P) Conditional derivation, 2 - 3�5. ((P → Q) → (~Q → ~P)) Conditional Derivation, 1-4
A Theorem
Homework
DeLancey 6.5 #2, 3a-c
Today’s animal: Aye-aye
Free Day
Homework
Today’s animal: Kiwi bird