Introducing PL trees
16.3 Signed and unsigned trees
16.3 Signed and unsigned trees
Let’s summarize again. We can test an argument for tautological validity by assuming that the inference is invalid and then ‘working backwards’, trying to find a consistent valuation which warrants that assumption. At various points, we may have to consider alternative valuations. For example, if we have arrived at (A∨B)⇒ T, then we have to consider the alternatives A ⇒ T and B ⇒ T. If,
twist and turn as we might, all such alternatives lead in the end to contradictory assignments of values to atoms, then we know that the original assumption that the inference is invalid is incorrect. In other words, if all branches of the ‘tree’ on which we set out the working can be closed off with absurdity markers, then the argument is valid. Conversely, if at least one alternative route through the tree involves no contradiction (even when all the complex wffs are fully unpacked), then the assumption that the argument is invalid is vindicated.
Now, so far, our ‘working backwards’ arguments we’ve been illustrating have been set out in the form of signed trees (we’ll speak of ‘trees’ here even if there is – as in §1 – only a trunk without any branching). The trees are ‘signed’ in the sense that the wffs that appear on them are all explicitly assigned a truth-value, either T or F. In the interests of neatness and economy, we are now going to move to using unsigned trees.
As a major step towards doing this, note that by a couple of simple tricks, we can easily ensure that every wff which appears on a ‘working backwards’ tree in fact gets the uniform value T. Here’s how:
• Trick one Instead of starting a tree by assuming that the premisses of the
argument to be tested are true and the conclusion false, assume that the premisses and the negation of the conclusion are all true.
• Trick two Thereafter, avoid ever writing something of the form ‘C⇒ F’
by always jumping instead to something of the form ‘¬C ⇒ T’. So, for example, when we meet on a tree a claim of the kind
(a) ¬(A∨B)⇒ T
don’t proceed via the obvious consequence
(b) (A∨B)⇒ F
to
(c) A⇒ F
(d) B⇒ F
but instead (trick two) jump straight from (a) to
(c*) ¬A ⇒ T
(d*) ¬B ⇒ T
And so on. The resulting ‘T’-only trees are only very slightly less natural than the ‘T-and-F’ trees we’ve been constructing so far.
Let’s quickly illustrate by re-doing a couple of examples in this style, with ‘T’- only trees: the changes required are marginal. So return first to
152 Introducing PL trees
D ¬(¬P∨Q),¬(¬R∨S)∴ (R∨¬P)
Starting this time by assigning truth to the premisses and the negation of the conclusion, and we have:
(1) ¬(¬P∨Q)⇒ T
(2) ¬(¬R∨S)⇒ T
(3) ¬(R∨¬P) ⇒ T
Now, let’s twice apply the rule just given for negated disjunctions (‘jump from
¬(A∨B)⇒ T to add both ¬A ⇒ T and ¬B ⇒ T’), and we immediately get
(4) ¬¬R ⇒ T (from 2)
(5) ¬S ⇒ T (from 2)
(6) ¬R ⇒ T (from 3)
(7) ¬¬P ⇒ T (from 3)
✼
We can close off the tree with an absurdity marker because of (6) and (4): no wff and its negation can be true together on a valuation. Since the tree starting from the supposition that the premisses and negation of the conclusion are true leads to contradiction, the argument must be valid. (And why are there two negation signs at line (4)? Because our rule said that if a negated disjunction is true, then each negated disjunct is true; and if a disjunct already starts with a negation sign, as in ‘¬R’ in (2), then it acquires another when negated.)
For a second ‘T’-only tree, return to
G ¬(P∧Q), (P∧R)∴ ¬(R∨Q)
Starting with the assumption that the premisses and negated conclusion are true we get: (1) ¬(P∧Q)⇒ T (2) (P∧R)⇒ T (3) ¬¬(R∨Q)⇒ T (4) P⇒ T (from 2) (5) R⇒ T (from 2) (6) (R∨Q)⇒ T (from 3) (7) ¬P ⇒ T ¬Q ⇒ T (from 1) (8) R⇒ T Q⇒ T (from 6)
The rule applied at (7) relies on the fact that, if a negated conjunction is true, then one or other of the conjuncts must be false (so we have the rule ‘given
¬(A∧B)⇒ T, consider alternatives ¬A ⇒ T and ¬B ⇒ T’). Not all branches of
this resulting tree lead to absurdity. So there is a coherent way of making the premisses and negated conclusion all true together, and so argument G must be tautologically invalid.
✼
16.4 More examples 153
It is worth quickly checking that our other sample trees can be rewritten with minimal changes as ‘T’-only trees. But why care that we can do this?
Because if we make assignments of truth standard on trees, and avoid ever assigning falsehood, then it isn’t really necessary to keep on writing down ‘⇒ T’ against each and every wff in our back-tracking reasoning. We might as well just let it be understood as ever-present. So from now on, let’s save time and ink by
using ‘T’-only trees and then keep the uniform assignments of truth tacit.
Suppressing the explicit assignments leads to unsigned trees. We’ll give some more examples in the next section.
16.4 More examples
Consider next the inference
H (P∨Q),¬(P∧¬R) ∴ (Q∨R)
To test for validity, we start from the assumption that the premisses and the negation of the conclusion are all true: so we start off our unsigned tree
(1) (P∨Q)
(2) ¬(P∧¬R)
(3) ¬(Q∨R)
with each line ending with a now invisible ‘⇒ T’. Since, as we said before, a negated disjunction is true just so long as both negated disjuncts are true, the assumption that (3) is true leads to the requirement that each of ‘¬Q’ and ‘¬R’ is true. So we add
(4) ¬Q
(5) ¬R
each with their invisible ‘⇒ T’. Now apply the principle that a disjunction is true so long as one or other disjunct is true, to get
(6) P Q (from 1)
✼
The alternative on the right-hand branch immediately leads to contradiction, for we have the occurrence of both ‘Q’ [⇒ T] and ‘¬Q’ [⇒ T] on that fork. So we close off that fork with the absurdity marker: no consistent valuation can be built on this branch. We’ve now extracted the juice from (1) and (3), so that leaves (2) to deal with. The relevant principle to invoke here is again that the negation of a conjunction is true when at least one of the negated conjuncts is true. So if ‘¬(P∧¬R)’ is true, one of ‘¬P’ or ‘¬¬R’ must be true. Applying that
to our tree, we get branching possibilities, so the full tree will look like this:
(1) (P∨Q)
(2) ¬(P∧¬R)
(3) ¬(Q∨R)
154 Introducing PL trees
(5) ¬R (from 3)
(7) P Q (from 1)
(8) ¬P ¬¬R (from 2)
✼ ✼
This time, both new forks lead to contradiction (we can’t have both P⇒ T and
¬P ⇒ T, as the left-most fork requires; nor can we have ¬R ⇒ T and ¬¬R ⇒ T
as the other fork requires). So all forking paths are closed off by contradictions. That means that there is no route which discloses a consistent valuation of atoms which makes (1), (2) and (3) all take the value T. Every alternative way of trying to make good the supposition that the inference in question is invalid runs into contradiction. So the inference is tautologically valid after all.
In constructing this last proof of validity, note that we chose first to draw con- sequences from (3) before making use of (1) and (2). There was no necessity about this; and we could have chosen to proceed in a different order, e.g. as follows: (1) (P∨Q) (2) ¬(P∧¬R) (3) ¬(Q∨R) (7) P Q (from 1) (8) ¬P ¬¬R ¬P ¬¬R (from 2) (9) ✼ ¬Q ¬Q ¬Q (from 3) (10) ¬R ¬R ¬R (from 3) ✼ ✼ ✼
Here we have proceeded more systematically, unpacking the complex wffs at the top of the tree in turn. Starting with (1), we asked what is required for that to be true, which immediately forces us to consider branching alternatives. Then fol- lowing up each of these alternatives, we considered the implications of (2) for
each of these alternative ways of proceeding; so we had to add the forking alter-
natives ‘¬P’ [⇒ T] and ‘¬¬R’ [⇒ T] to both existing branches. So we now have
four options to consider. We can close off one path because it contains contra-
dictory wffs which can’t both be true. Then we proceeded to extract the implica- tions of (3) for the three options that remain open. For (3) to be true, both ‘Q’ and ‘R’ must be false: so we add ‘¬Q’ [⇒ T] and ‘¬R’ [⇒ T] to every branch that still remains open. We now end up with every route leading to absurdity, and so closed off with the absurdity marker. Hence, as we would hope, we arrive at the same result as before: H must be valid. (Later, in §19.3, we’ll formally prove that it can never make any difference in which precise order we unpack wffs when ‘working backwards’.)
Another example: what of the following inference? ✼
16.4 More examples 155
I (P∨ (Q∧ R)), (¬P∨R), ¬(Q∨¬¬S) ∴ (S∧R)
We start off in the now familiar way, with the premisses and negated conclusion:
(1) (P∨ (Q∧ R))
(2) (¬P∨R)
(3) ¬(Q∨¬¬S)
(4) ¬(S∧R)
A negated disjunction is true only if both disjuncts are false, i.e. the negated dis- juncts are true. So (3) implies a couple of definite truth assignments
(5) ¬Q (from 3)
(6) ¬¬¬S (from 3)
Where now? Well, we know that a pair of negation signs in effect cancel each other out. So we can infer
(7) ¬S (from 6)
Let’s now unpack the other complex wffs in turn. First we have
(8) P (Q∧ R) (from 1)
The right-hand branch doesn’t close off immediately, but if we have (Q∧ R) ⇒ T on that way of making (1) true, then we must also have Q⇒ T and R ⇒ T. So we can add to the right hand branch thus
(9r) Q (from 8)
(10r) R (from 8)
And we’ve hit a contradiction between (5) and (9r) and can close off that branch. Proceeding down the left-hand branch, we’ve so far ensured that (1) and (3) are true. To make (2) true we need either ¬P ⇒ T or R ⇒ T. So we have
(9l) ¬P R (from 2)
and the left-hand of the two branches we’ve added immediately closes off, since we can’t have ‘P’ and ‘¬P’ both true. That leaves us with only (4) to unpack. So we apply the rule for negated conjunctions again and the resulting tree from line (7) again looks like this:
(7) ¬S (from 6)
(8) P (Q∧ R) (from 1)
(9) ¬P R Q (from 2: from 8)
(11) ¬S ¬R ✼ (from 4)
✼
So, all but one of the branches close. Looking at that remaining branch,
✼ R
156 Introducing PL trees
however, and putting back the suppressed assignments of truth, we can read off the following valuation
P⇒ T, ¬Q ⇒ T (so Q ⇒ F), R ⇒ T, ¬S ⇒ T (so S ⇒ F).
This is designed to make the premisses of I true and the conclusion false (check that it does!). So the argument is tautologically invalid.
This tree technique is elegant and (usually) efficient. Our next task is to gather together the various principles we have found for tree-building, check whether we need any others, and put them together in one systematic list. Though first, let’s briefly pause for breath.
16.5 Summary
• Testing an argument for tautological validity is a matter of searching for ‘bad’ valuations. We can do this by a brute-force inspection of all possible valuations. Or we can ‘work backwards’, i.e. assume there is a bad valua- tion and try to determine what it would look like. If the supposition that there is such a bad valuation leads to contradiction, then the argument in question is valid.
• The ‘working backwards’ method, in the general case, will necessitate inspecting branching alternative possibilities, e.g. looking at alternative ways of making a disjunction true. The resulting working is naturally laid out in the form of a downward branching tree.
• By some simple tricks, we can ensure that every wff on such a tree is assigned the value T. But then, if every wff is assigned the same value, we needn’t bother to explicitly write down the value – giving us ‘unsigned’ trees.
• The official rules for tree-building follow in the next chapter.
Exercises 16
Although we haven’t yet laid down formal rules, it is worth doing a bit of infor- mal exploration to get more feeling for the tree method before we dive into the official story. So, see how you get on testing the following inferences for validity, using the techniques of this chapter. At this stage it might help to keep writing down explicit ‘T’-only trees, appending ‘⇒ T’ alongside every wff to remind yourself what you are doing and why you are doing it.
1. (R∨P), (R∨Q)∴ (P∨Q) 2. ¬(P∧¬Q), P ∴ Q 3. (¬P∨Q), (R∨P)∴ (R∧Q) 4. ((P∧¬Q)∨R), (Q∨S)∴ (R∧S) 5. P,¬(P∧¬Q), ¬(Q∧¬R), (¬R∨S)∴ S 6. ((P∧Q)∨ (R∧P))∴ ((Q∨R)∨P)
This chapter gives a more formal description of the tree test for tautological validity. The first section, then, is necessarily going to be rather abstract. But it will really do no more than carefully formulate the principles we’ve already used in the last chapter. (Those who get a bit fazed by this sort of abstraction can try skimming through the statement of the rules and the following illustrative sec- tions a few times in quick succession, using each to throw light on the other.)