## Do (A implies B) and (A implies notB) contradict each other?

• 1.8k
Do (A implies B) and (A implies notB) contradict each other?

Please give your reasoning, if you choose to answer, for why you think they do, or don't, contradict each other. And if you think they do contradict each other, does that mean they can't both be true at the same time?
1. Do (A implies B) and (A implies notB) contradict each other? (23 votes)
Yes
39%
No
61%
• 2.2k
Two propositions contradict if the truth or falsity of one entails the falsity or truth of the other (i.e. (P ↔ ~Q) ^ (~P ↔ Q))*. Formally, then, they contradict if it is true that:

((A → B) ↔ ~(A → ~B)) ^ (~(A → B) ↔ (A → ~B))

...Which turns out to be false (link). Given material implication, they do not contradict when A is false, and therefore the two are not full-on contradictory. There would only arise a contradiction supposing A.

(NB: Given the way that common speech differs from material implication, in common speech the two speakers would generally be contradicting one another.)

* I think the second conjunct is redundant, but it illustrates the principle of contradiction.
• 4.4k
A contradiction is of the form "P ^ ~P"

"A implies B" =/= "A implies not-B", and so the conjunction is "P ^ Q" rather than "P ^ ~P" -- the contradiction would be "A implies B" and "A does not imply B" -- the "not" would have to apply to the logical connective "imply".

Implication (EDIT: i.e. "A implies B") is logically equivalent to "~A or B", so...

Let
P = A implies B
Q = A implies not-B

then...
P = not-A or B
Q = not-A or not B

P =/= Q, and therefore you cannot derive the form "P ^ ~P", and so they do not contradict.

Combining P and Q what we instead obtain is: not A or B or not B (EDIT: or, we should say, not-A or B and not-A or not B), and since "B or not B" is a tautology we can simplify the expression to "not-A" -- the "B" is a sideshow because it doesn't matter if it's true or false when we put P and Q in conjunction, all that matters ifis the truth value of A.
• 222
I voted that "(A implies B) and (A implies notB)" do not contradict each other. But the only possibility is if A is false/falsy.

Here is a JavaScript example:

// Define the values of A and B
const A = false; // A must be false for both implications to be true
const B = true; // B can be any value, but it doesn't matter because A is false

// Logical implication function
function implies(p, q) {
return !p || q;
}

// Check the implications
const A_implies_B = implies(A, B); // A implies B
const A_implies_notB = implies(A, !B); // A implies not B

// Output the results
console.log(A: ${A}); console.log(B:${B});
console.log(A implies B: ${A_implies_B}); console.log(A implies not B:${A_implies_notB});

// Check if both implications are true
const result = A_implies_B && A_implies_notB;
console.log((A implies B) and (A implies not B): \${result});


In the example above the following occurs:

• The implies function takes two arguments, p and q, and returns true if p implies q (which is equivalent to !p || q).
• We define the values of A and B such that A is false.
• We check the implications A implies B and A implies not B.
• Finally, we print the results and check if both implications are true.

When we run this code, we see that both implications are true, demonstrating the logical conditions.
• 2.2k
A contradiction is of the form "P ^ ~P"

Presumably you could also run a truth table on this form. If neither conjunct is inherently fallacious and yet the truth table comes out fallacious (i.e. the composite proposition is always false), then presumably there would be a contradiction. The composite proposition would be:

(A → B) ^ (A → ~B)
• 4.4k
Oh definitely! That's how I'd have preferred to set it out, but then I saw it was just easier to type it out :D -- but I like truth-tables because they just show it all laid out. They're mathematically clunky but conceptually useful for showing the structures for learning.

And, yes, I agree with your rendition of the composite proposition. I just find it easier to think of material implication in terms of the equivalence between: A -> B <-> not A or B (Who said that philosophers never agree on things? As long as we stipulate the definitions... :D )
• 3.2k
If A is false, then (A->B) & (A->~B) is true. So (A->B) & (A->~B) does not entail a contradiction. It's that simple.
• 2.2k
- :up: ...and now a forgotten pm comes to mind. :blush:
• 6.1k
A=>B & A => !B

A true and B true means... The thing's false
A true and B false means... The thing's false
A false and B true means... The thing's true.
A false and B false means... The thing's true.

So no, it's not contradictory. As if A is false the statement is true, since both implications are true.

A & A=>B & A=> !B

A true and B true means... The thing's false.
A true and B false means... The thing's false.
A false and B true means... the thing's false.
A false and B false means... the thing's false.

I think where you're getting the contradiction from is the thought that stipulating A lets you derive B and !B, if you also stipulated A=>B and A=>!B. But in fact that's stipulating A and A=>B and A=>!B together.
• 3.5k
False implies anything, so (by example) nope.
• 2.5k
We have to be careful by what is meant by 'implies' In logic, the word, necessarily is used more often in these examples. "Necessarily leads to B" So if you have A -> B and A -> !B, that is a contradiction.

Now, if you want to use English instead of logic, we can also examine that.

Lets use the word 'implies" which for many means, "could" or "not necessarily".

In which case A could, or could not lead to B. In other words, you're describing an induction. We cannot use A -> B here either, as that's not what it means. Once we had the result of the induction, there are two results we can conclude.

A always leads to (outcome) or

This can be determined through testing. For example, if I flip a coin, it sometimes comes out heads, and sometimes comes out tails. This can be written in logic as "Some A's lead to Bs, and some A's lead to not B's"

If of course The question was whether a coin always spins when flipped, we would see it would flip every time. In this case after the test it would be A -> B.

• 1.8k
What do you think of this?

Do you think Uncle Joe's argument is valid?
• 2.6k
It is troublesome to talk about these things if one is not using very specific terminology. What does "contradictory" really mean? As Leontiskos has shown, the two formulas yield the same result if A is false (true for any value of B).

A contradiction is an assertion of Propositional Logic that is false in all situations; that is, it is false for all possible values of its variables.

If (A implies B) and (A implies notB) is understood as (A → B) ^ (A → ¬B), it is not contradictory as it is true whenever A is false.

I think what is being asked here is whether one is the denial of the other. And the answer is no. Putting it in logical tables, denial would be whenever (A → B) yields True (A → ¬B) yields False. We know that there are values for A and B where both yield the same result.

The denial of (A → B) would be ¬(A → B), same for the other (p/¬p). A less obvious example is that ((A∧¬B) ∨ (¬A∧B)) is the denial of ((A∧B) ∨ (¬A∧¬B)). The first is the XOR gate, the second is XNOR. For any value of A and B, one will give False and other True.

The issue is, whenever A is true, the two do not give the same results. Meaning, if A is true (it is raining), it can't be true that A implies B (it is wet) and not-B (it is not wet) at the same time. As fdrake said.

But if we say it is not raining, the antecedent (A) is false, and the way material implication works in classical logic is that, if the antecedent is false, the implication is always true — I remember we both took part in an excruciatingly long discussion about denying the antecedent.

Hoping my explanation was clear and free of errors.
• 281
I am not a logician, but might...
"A -> not A"
mean that
"not (A -> A)."
But surely A -> A. Therefore, not "A -> not A."
• 1.8k
what does that have to do with this stuff?
• 281
That is the argument you forwarded.
• 281
A -> not A
• 1.8k
I don't see the connection between what I said and what you're saying
• 281
Folks, have rightly been objecting to A in my opinion. But if you assume B or not-B, you end up with Not-A.

Thus, A -> (notB or B) -> not-A. Or more succinctly, A -> not-A
• 1.8k
it's not

A -> (notB or B)

It's more like

A -> (notB and B)

But yes, it makes sense why that would imply not a. I think your Comms in this thread have just been a little too succinct, you're not filling in enough blanks for people to follow your logic, what point you're making, or even if you answered yes or no to the op poll question
• 281
• 2.6k

TL;DR: Denial (¬p) and contradiction (yields false in all cases) are not the same thing. Denial is a relationship between two propositions, contradiction is a feature of a single proposition. The two propositions given are not the denial of each other, so Prop1&Prop2 is not contradictory (it can yield True for some values). Material implication (→) is defined in such a way that when the antecedent is False, the result is always True. So, if the antecedent is False, both propositions are True.
• 4.4k
I appreciated the Truth-Tables.

I like this post because it's getting into why we might be tempted to say they contradict.

I think, at least in philosophy though maybe there's some other argument this stems from that I'm not aware of, that we should separate out implication from modality -- so when you introduce "possibility" and "necessity" those are entirely different operators from implication.

Though if there's some other argument going on I'm not aware of then you can link it -- I'm surprised to find so many people saying "Yes". lol
• 2.4k
Can anyone think up a real world example where you would point out that A implies both B and not-B except for saying something along the lines of:

"A implies B and not-B, therefore clearly not-A."
• 13k
Can anyone think up a real world example where you would point out that A implies both B and not-B except for saying something along the lines of:

"A implies B and not-B, therefore clearly not-A."

Fuzzy logic/Many-valued logic...
• 2.4k

What would you be trying to say? I am having trouble thinking of an example.
• 13k

Thinking in terms of a Venn diagram, you could have many points of congruence or intersection between sets and be able to say something along the lines of what your previous post meant.
• 4.4k
Can anyone think up a real world example where you would point out that A implies both B and not-B except for saying something along the lines of:

"A implies B and not-B, therefore clearly not-A."

That's not the same as "(A implies B) and (A implies not-B)" -- that'd be "(A implies (B and not-B)).

A real world example is often hard to parse into material implication -- sometimes, yes, but sometimes it's hard -- the conjuncts of disjuncts, while they can be claimed, is even rarer :D

Though after we dismiss "B and not-B" as always false, we can see that the truth of the proposition will only rely upon A, since "implies" is logically equivalent to "not-A or (B and not-B)", and the truth of a disjunct is true if one of the propositions is true -- so if not-A is true then it is true, and if not then it is false -- since not all results in the truth-table are false it is not a contradiction.
• 13k
A real world example is often hard to parse into material implication -- sometimes, yes, but sometimes it's hard -- the conjuncts of disjuncts, while they can be claimed, is even rarer :D

Though after we dismiss "B and not-B" as always false, we can see that the truth of the proposition will only rely upon A, since "implies" is logically equivalent to "not-A or (B and not-B)", and the truth of a disjunct is true if one of the propositions is true -- so if not-A is true then it is true, and if not then it is false -- since not all results in the truth-table are false it is not a contradiction.

In epistemic logic, such cases are known as known known's, known unknown's, and unknown unknowns. Yet, there are examples of making false inferences like unknown known's that would describe such a situation arising in epistemic logic...
• 2.6k
That's not the same as "(A implies B) and (A implies not-B)" -- that'd be "(A implies (B and not-B)).

Maybe the confusion comes from taking A as a proposition instead of a variable that may be 0 or 1.
• 4.4k
Would that make a difference? 0/1=F/T as I understand it.
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