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Truth Trees

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Truth Trees

- The problem with standard truth tables is that they grow exponentially as the number of sentence letters grows, so…
- Most of our work is wasted because most of the Ts and Fs we plug in don’t show anything!
- Testing for consistency, for example, only the presence or absence of an all T row is relevant!
- But indirect truth table only work effectively for rigged examples
- We need something better: i.e. Truth Trees!

- Truth trees are an improved version of the short-cut truth table method for determining consistency and validity
- Both methods assign truth values to whole sentences and then figure out what truth values of their components produce the assigned truth value—we are, in effect, “decomposing” the sentences
- Both methods test to see whether it is possible to produce a correct truth value assignment to the sentence letters that gets the assigned truth value for the whole sentences
- Recall the short cut truth table test for consistency…
- \\

A B / B (C A) / C B / A

T

T

T

T

Write the sentences on one line with slashes between them

Assign true to each sentence by writing ‘T’ under its main connective

A B / B (C A) / C B / A

T

T

T

T

F

Since A is true, A must be false so this truth value is “forced” on A

Assign “forced” truth values.

We start with the last sentence because assigning true to the other sentences doesn’t “force” truth values on their parts.

A B / B (C A) / C B / A

F

F

T

T

T

T

F

Now that we’ve assigned a truth value to A, other truth values are forced by that:

All the other A’s must be false too!

A B / B (C A) / C B / A

T

T

F

T

F

T

T

T

T

F

This forces more truth values:

Since A is false, to make the first sentence true we have to assign true to B—which makes all the B’s true.

A B / B (C A) / C B / A

T

T

F

T

F

T

T

T

F

T

F

Since B is true, B must be false—so yet another truth value is forced

A B / B (C A) / C B / A

T

T

F

T

F

F

T

T

T

F

T

F

F

Since B is false, C must be false in order to make the conditional, C B, true--so we have another forced truth value: all C’s have to be false

A B / B (C A) / C B / A

T

T

F

T

F

F

T

T

T

F

T

F

F

F

Now we can complete the truth value assignment—and there’s only one way to do it: by assigning false to C A, since both of its parts are false.

A B / B (C A) / C B / A

T

T

F

T

F

F

T

T

T

F

T

F

F

F

But this isn’t a possible truth value assignment because it says that the conditional,B (C A), is true even though its antecedent is true and its consequent false.

And there’s no way to avoid this since all truth values were forced!

A B / B (C A) / C B / A

T

T

F

T

F

F

T

T

T

F

T

F

F

F

This shows that there’s no truth value assignment that makes all sentences true

Therefore that this set of sentences is inconsistent.

A B / B (C A) / C B / A

T

T

F

T

F

T

T

T

T

F

T

F

T

T

Note: if you assigned truth values in a different order the problem will pop up in a different place (see Hurley p. 40)—but it will pop up somewhere, like a lump under the carpet!

- In doing a truth tree we start in the same way: by assigning truth values to whole sentences and then working backward until we’ve assigned truth values to all sentence letters.
- We do this by “growing” a tree-structure according to tree rules which “decompose” sentences into their constituent sentence letters.
- The tree rules represent the ways in which the sentence forms to which they apply are made true—so, e.g.
p q is made true by either p’s being true or q’s being true

(p q) is made true by p being true and q being true

- p q is made true by either p’s being true or q’s being true
- In both cases where p is true, p v q is true.
- In both cases where q is true, p v q is true.

- (p q) is made true by p’s being true and q’s being true
- That is, by both p and q being false
- We can construct tree rules from the characteristic truth tables for the connectives in this way!

To represent the truth value assignment that makes a sentence true we want to show truth flowing up the tree—like sap from the roots

Except in this case truth flows upward from the branches!

T

T

T

DN

p

p

- The rule for Double Negation is: rewrite, erasing two ’s
- Sentences that are basically OR’s are represented as branching structures
- Sentences that are basically AND’s are represented by non-branching structures.
- We understand conditionals and biconditionals as basically OR’s and AND’s

OR

p q

p q

AND

p • q

p

q

p • qpq

p v q

p q

To make p v q true, all we need is truth flowing through one of its parts

So we represent disjunction by a branching rule

Truth

Truth

- To make p • q true, Truth has to flow through both p and q
- So we represent conjunction by a non-branching rule

Truth

- Conditional and biconditional are actually “extras” in our language: we can say everything they say just in terms of conjunction, disjunction and negation.
- p q is equivalent to either p OR q so we formulate the tree rule for conditional as a branching OR rule
- p q is equivalent to either (p AND q) OR ( p AND q) so we formulate the tree rule for biconditional as a branching OR rule with ANDs on both branches.
- To see why this is so, consider the truth tables for conditional and for biconditional

- p q is true if either p is false or q is true so it’s logically equivalent to p v q
- You can prove this by testing the two sentences for equivalence!

- p q is true if either both p and q are true or both p and q are false.
- So it’s equivalent to (p • q) v ( p • q)
- Note: we’re helping ourselves to the idea that saying p is false is the same thing as saying p—which is ok given the truth table for

p qp pq q

p q

p q

p q says either p • q or p • q so it’s a branching rule with conjunctions on both branches

Truth has to either flow through both p and q or through both p and q

Truth

Truth

- To make p q true, Truth has to flow through either p or q
- p says p is false so this says what makes p q is p being false or q being true

Truth

Truth

(p q) p p q q

(p q)p q

(p q) says that p and q have opposite truth value

Truth has to either flow through p and q or through p and q

- (p q) says p q is false
- What makes a conditional false is true antecedent, false consequent
- So we represent this as a conjunction of p and q

Truth

Truth

Truth

(p v q) p q

(p • q)

p q

(p • q) is equivalent to p v q by DeMorgan’s Law

So we represent (p • q) by this branching rule

Truth

Truth

- (p v q) is equivalent to p • q by DeMorgan’s Law
- So we represent (p q) by this non-branching rule

Truth

pp

- The double negation rule is obvious!
- p is equivalent to p so, a fortiori, p makes p true.

Truth

- We use the rules to grow the tree downward.
- We apply the tree rules to each sentence successively to “decompose” it into simpler sentences that make it true…
- …and we decompose those sentences into even simpler sentences…
- …until we get down to sentences that can’t be decomposed any further, that is
- Sentence letters and negations of sentenceletters
- Then the tree is complete.

P Q

P • Q

Q

- Write the sentences to be tested in a vertical column: these are the initial sentences
- We’re looking for a truth value assignment that will make all of them true (if there is one)
- So we start by considering truth value assignments that make each of them true individually
- And see if we can put them together

- Apply tree rules to each sentence to which they apply, checking sentences when they’ve had rules applied to them
- We start with non-branching rules to keep the tree from getting too big.

√

~ P

Q

√

- Now we apply the rule for conditional to P Q writing the result at the bottom the tree
- The tree stops growing because no further rules can be applied.

√

~ P

Q

~ P

Q

√

- A “branch” or “path” is the result of tracing from each sentence at the bottom of the tree all the way up to the top
- There are 2 (overlapping) branches on this tree: the initial sentences are on both branches.

√

~ P

Q

~ P

Q

√

- Each branch wants to represent a truth value assignment to the initial sentences which we can read off as follows:
- If a sentence letter occurs on a branch, TRUE is assigned to that sentence letter; if the negation of a sentence letter occurs, FALSE is assigned to that sentence letter.

√

~ P

Q

~ P

Q

√

- On this tree, both branches assign FALSE to P and TRUE to Q
- So each branch represents the same truth value assignment, viz.
- The truth value assignment represented by the row of the truth table in which all sentences got true, remember…

√

~ P

Q

~ P

Q

P Q / P • Q / Q

T

T

T

F

T

F

T

T

T

F

F

F

T

F

F

F

P is FALSE;Q is TRUE

F

T

T

T

F

T

T

T

F

T

F

T

F

F

F

F

Consistent or inconsistent?Consistent

We constructed this row of the truth table on the truth tree without wasting time doing the other rows that didn’t matter!

- The left branch doesn’t represent a truth value assignment because it assigns both TRUE and FALSE to P!
- So we say that branch is “closed” and indicate that by putting an X at the bottom

√ P Q

√ P • Q

P

Q

~ P

Q

X

- A completed tree is open if it has at least one open branch.
- A completed tree is closed if it has no open branches, i.e. if all of its branches are closed.
- Consistency only requires the some (i.e. at least one) truth value assignment make all the sentences true so
- If the tree is open, then the initial sentences are consistent
- If the tree is closed, then the initial sentences are inconsistent

- So now we can do two things:
- We can determine whether a set of sentences is consistent or inconsistent
- Open tree – consistent
- Closed tree – inconsistent

- And if the sentences are consistent we can determine which truth value assignment(s) makes them all true by reading the the open branch(es)

- We can determine whether a set of sentences is consistent or inconsistent
- But what if a set of sentences is inconsistent?

- This tree is closed so the initial sentences are inconsistent.
- There is no truth value assignment that makes all initial sentences true.

InitialSentences

√

P Q

P ~Q

~ P

Q

X

X

- Know the tree rules and how how they are derived
- Be able to invent a tree rule for a symbol if given its characteristic truth table
- Grow a truth tree
- Determine what a completed truth tree tells you about the consistency or inconsistency of initial sentences
- If the initial sentences are consistent, determine which truth value assignment makes them all true
- Given a completed tree, determine what its initial sentences are.

(P Q) R

- Write out the argument vertically, premises first and then conclusion
- The truth tree test for validity is an indirect proof method (aka reductio, proof by contradiction): we want to show that it’s not possible for all the premises to be true and the conclusion false.
- So we ask: “What if the premises were true and the conclusion were false?”

R

P

(P Q) R

- To ask that question, we negate the conclusion, grow a tree, and see what happens.
- When we test an argument for validity, we call the premises + the negation of the conclusion, the sentences above, the initial sentences.
- We then test these initial sentences for consistency by growing a truth tree from them.

R

negation of the conclusion

P

(P Q) R

- We know that:
- If the premises + negation of conclusion are consistent the argument is invalid.
- If the premises + negation of conclusion are inconsistent the argument is valid.

- So by testing these sentences for consistency, we can determine whether the argument is valid or invalid!

R

negation of the conclusion

P

P1 / P2 / . . . Pn / ~ C

T T T T

When we say that the premises + the negation of the conclusion are consistent we’re saying that there’s a truth value assignment (row of truth table) in which all these sentences are true.

Please run this by me again

P1 / P2 / . . . Pn // ~C

T T T T F

If there’s a row in which all the premises and the negation of the conclusion are true then in that very row all the premises are true and the conclusion itself is false.

So the argument is invalid!

Please run this by me again

P1 / P2 / . . . Pn / ~ C

T T T T

Inconsistent: there’s no row like this

Now suppose that the premises + the negation of the conclusion are inconsistent.

This means that there’s no row in which the premises and the negation of the conclusion are all true.

Please run this by me again

P1 / P2 / . . . Pn // ~C

T T T T F

Valid: there’s no row like this

So there’s no row in which all the premises are true and the conclusion itself is false.

So the argument is valid!

Please run this by me again

- Using the tree method, we test for validity by testing the initial sentences—premises + negation of conclusion for consistency.
- If the initial sentences are consistent the argument is invalid.
- If the initial sentences are inconsistent the argument is valid.
- So now let’s try it!

(P Q) R

- We’re going to test these initial sentences for consistency.
- If the tree closes, they’re inconsistent, so the argument is valid.
- If the tree is open, they’re consistent, so the argument is invalid.

R

negation of the conclusion

P

(P Q) R

We apply the double negation rule to this sentence, check it, and write the result at the bottom of the tree

R

√ P

P

√(P Q) R

We apply the rule for conditional to this sentence, check it, and write the result at the bottom of the tree

R

√ P

P

(P Q)

R

Are there any problems? We check both branches to see whether either of them includes a sentence and its negation.

Note: a sentence is “included” on a branch if it occurs on a line by itself—not just as part of a longer sentence.

√(P Q) R

R

√ P

P

(P Q)

R

X

We’ve got a problem: R and R are on the same branch, so that branch stops growing and closes.

We show that the branch is closed by putting an X at the bottom.

√(P Q) R

R

√ P

P

Now we apply the negation of a disjunction rule to this sentence

√ (P Q)

R

X

- P
- Q

The tree is now finished growing because each sentence to which a rule could be applied has been checked—showing that the appropriate rule has been applied to it.

Is there a problem?

√(P Q) R

R

√ P

P

√ (P Q)

R

X

- P
- Q

X

Yes! The remaining branch includes P and P so it closes, and we show that by putting an X at the bottom of the branch.

The tree is now complete and it is closed—so the argument is valid!

√(P •Q) R

R

√ P

P

√ (P •Q)

R

X

P

Q

This tree has finished growing but is open so the argument is invalid.

We can also determine some more things about this argument by “reading” its truth tree…

√(P •Q) R

- Reading from the bottom up, we look for the first sentence which wasn’t the result of applying a tree rule.
- That sentence is the negation of the conclusion.
- So the conclusion of this argument is P

R

√ P

P

√ (P •Q)

R

X

P

Q

√(P •Q) R

- The initial sentences (the premises + negation of the conclusion of the argument) are consistent.
- The open path represents a truth value assignment that makes all the initial sentences true.

R

√ P

P

√ (P •Q)

R

X

P

Q

√(P •Q) R

- If a sentence letter appears on an open path, that truth value assignment assigns TRUE to that sentence letter.
- If the negation of a sentence letter appears, it assigns FALSE to that sentence letter

R

√ P

P

√ (P •Q)

R

X

P

Q

√(P •Q) R

- So, the truth value assignment that makes all initial sentences true is…
- P – TRUE; Q – TRUE; R – FALSE

R

√ P

P

√ (P •Q)

R

X

P

Q