# What is a proposition?

An assertion is a statement.

A proposition is an assertion which is either true or false (but not both).

The Following are propositions:

• 4 is a prime number
• 3 + 3 = 6
• The moon is made of cheese.

The following are not propositions:

• x+y > 4.  Is this true or false? It depends on the value of x and y, the statement takes a true or false value.
• x=3. You cannot associate a truth value to this because it simply assigns a value to x.
• Are you leaving? This is not an assertion, it is a question.
• Buy 4 Books This is not an assertion, it is an order.

# English to Logic

Formalize the following arguments as syntactic sequents of the propositional calculus,
giving an explicit interpretation of your sentence-letters.

The murderer was either Colonel Mustard or Professor Plum.  But it wasn’t
Professor Plum.  So it was Colonel Mustard.

P = The murderer was Colonel Mustard
Q = The murderer was Professor Plum

(P x Q) ^ not Q => P

We can see from the table that (P x Q) ^ not Q => P is a logical implication (tautology) and that we can conclude that P must be true if P x Q ^ not P is true. Therefore it is proved that The murderer was Colonel Mustard.

Either the Master or the Dean was in the library.  But if the Master wasn’t there,
the Dean wasn’t there either.  So they were both in the library.

P = the Master was in the library.
Q = the Dean was in the library.

not (P x Q) => (P ^ Q)

We can see from the table that not (p x q) does imply (p ^ q) and that p and q are true, thus it is proven that the Master was in the library and the Dean was in the library.

You can only buy a Young Persons railcard if you’re under 26 or a student;
otherwise not.  If you can buy a Young Persons railcard, you can get discounted
train tickets.  But you’re not under 26.  So unless you’re a student, you can’t get
discounted train tickets

P = You are under 26
Q = You are a student
R = You can buy a Young person rail card
S = You can get a discount

not (((P v Q) => R) => S) => not S v Q

(I cannot be bothered to do a truth table for this – so  cannot prove that my answer is correct – but I am pretty sure that is – so mybe you would like to do one and let me know?)

If God is willing to prevent suffering, but unable to do so, He is not omnipotent. If
He is able to prevent suffering, but unwilling to do so, He is not loving. If God
exists, He is loving and omnipotent. And if He is both willing and able to prevent
suffering, then there can’t be any suffering – but there is. So God doesn’t exist.

P = God is willing to prevent suffering
Q = God is omnipotent
R = God is able to prevent suffering
S = God is loving
T = God exists
U = There is no suffering

not (((P ^ not R) => not Q) ^ ((P ^ not P) => not S) ^ ((P ^ R) => not U)) => not T

(I cannot be bothered to do a truth table for this – so  cannot prove that my answer is correct – but I am pretty sure that is – so maybe you would like to do one and let me know?)

The protesters will go away if Oxford stops experiments on animals. But this
could only happen with government intervention. So, unless the government
intervenes, they won’t go away

P = The protesters will go away
Q = Oxford stops experiments on animals
R = the government intervenes

not ((Q => P) => R) => not P

Here we can see that not ((Q => P) => R) => not P , not P and not ((Q => P) => R) are all true, thus we can conclude that not ((Q => P) => R) => not P is true.

# 3 no namers – logical implications

(P=>Q) => ((Q => R) =>(P => R))

This is actually the hypothetical syllogism in another form. For by considering (P Q) as a proposition S, (Q R) as a proposition T, and (P R) as a proposition U in the hypothetical syllogism above, and then by applying the “exportation” from the identities, this is obtained

((P => Q) ^ (R => S)) => ((P ^R) => (Q ^ S))

For example, if the statements “If the wind blows hard, the beach erodes.” and “If it rains heavily, the streets get flooded.” are true, then the statement “If the wind blows hard and it rains heavily, then the beach erodes and the streets get flooded.” is also true.

((P <=> Q) ^( Q <=> R)) => (P <=> R) This is saying that you can infer that P is equal to R when ((P <=> Q) ^( Q <=> R)).

This just says that the logical equivalence is transitive, that is, if P and Q are equivalent, and if Q and R are also equivalent, then P and R are equivalent.

# Hypothetical Syllogism – a logical implication

((P=>Q) ^ (Q => R)) => (P => R) named Hypothetical Syllogism

P => Q (if P imples Q and…)
Q => R (if Q implies R…)
Then, P must imply R.

P = I do not wake up
Q = I cannot go to work.
R = I will not get paid.

If I do not wake up, then I cannot go to work. (P => Q)
If I cannot go to work, then I will not get paid. (Q => R)
Therefore, if I do not wake up, then I will not get paid. (P => R)

# Disjunctive Syllogism – a logical implication

(not P ^ (P v Q) => Q named Disjunctive Syllogism formerly known as modus tollendo ponens meaning if not P and P or Q then we can conclude Q.

P v Q (if P or Q is true and….
not P (is true)
then Q must be true.

if P = The car is fast
and Q = The car comes in first

Then the compound statement: “If the car is not fast and (the car is fast or the car comes in first) then the car comes in first” is always true.

Suppose that (P) is false, the car is fast
Suppose that (Q) is true, the car comes in first

Then if the car is not fast and (the car is fast or the car comes in first) then we can conclude that the car comes in first. What happened was that the statement the car is not fast removes the statement the car is fast – all that is left is the “the car comes in first” – if that statement is true then the conclusion can only be that it is true that that the car comes in first.

So you might say as a human, wow the car was not first, but it came in first.

# Simplification – a logical implication

(P ^ Q) => P named Simplification meaning if (P ^ Q) then P
(P ^ Q) => Q named Simplification meaning if (P ^ Q) then P

If the compound statement  (P ^ Q) => P is given and is true (it is always true) and P ^ Q is true then P must be true. (see truth table to follow the reasoning).

The same reasoning applies to (P ^ Q) => Q.

P = I cannot get hold of any money
Q = The bank will not lend me any money

The the compound statement: “if I cannot get hold of any money and the bank will not lend me any money then I cannot get hold of any money” is always true.

Suppose that, it is true that (P) I cannot get hold of any money
Suppose that, it is true that (Q) The bank will not lend me any money

Then we can conclude that it is true that: “I cannot get hold of any money – if I cannot get hold of any money and the bank will not lend me any money”. You can see that just more truth (The bank will not lend me money) to some original truth (I cannot get hold of any money) – So we can take it that the original truth is still true if both the original truth (I cannot get hold of any money) and the added truth (the bank will not lend me any money) are both true.

# Addition – a logical implication

P => (P v Q) meaning, P implies (P or Q)

If the compound statement P=>(P v Q) is given and is true (it is always true) then we can conclude P by just taking P.

If P = the traffic light is green
and Q = the traffic light is red.

Then in English, P=>(P v Q ) will be “If the traffic light is green then the traffic light is green or (otherwise) the traffic light is red”. As mentioned, this compound statement is always true.

So, If the compound expression (P => (P v Q)) is given, we know it is true, so we can always conclude P by just taking P.

Suppose that, it is false that (P) the traffic light is green,
Suppose that, it is true that (Q) the traffic light is red,

Then we can conclude that it is false that the traffic light is green, because P is false.

Suppose that, it is true that (P) the traffic light is green,
Suppose that, it is false that (Q) the traffic light is red,

Then we can conclude that it is true that the traffic light is green, because P is true.

Suppose that it is true that (P) the traffic light is green,
Suppose that it is true that (Q) the traffic light is red,

Then normally we would conclude that the traffic light is broken, but in this case, logically we would conclude that the traffic light is green because P is true.

Suppose that it is true that (P) the traffic false is green,
Suppose that it is true that (Q) the traffic false is red,

Again, normally we would conclude that the traffic light is broken, but in this case, logically we would conclude that it is false that the traffic light is green because P is false.

In the truth table we can see the tautology in P=>(P v Q) – the compound statement is always true – which proves what we have said in the examples.