A tableau is a backward chaining proof method that seeks to demonstrate that no counter-model for a proposition exists, or to find one if it does. Tableaux start by asserting all the assumptions of an argument and the negation of the conclusion. Why the negation of the conclusion? The aim is to discover whether there is a non-contradictory way that the assumptions can be true while the conclusion is false. If so, then the inference is not valid (and the counter-model can be exhibited). If not, the inference must be valid.
There are many different systems and notations of tableaux. It is sometimes hard to render without diagrams, but I’ll try using a set-based notation. The system I will use is similar to those found in the literature; and it can be proven sound and complete with respect to the (modal) logic. The argument below is somewhat technical, so feel free to trust me and jump to my conclusion :).
Example: Not Tisthammer’s Argument
At each step, we’ll have a set of propositions that are true at a particular world. I will write A:i, where i is an integer identifying a world to mean “A is true at world i”. The initial set contains the assumptions and the negation of the conclusion, all true at some “starting world” (world 0).
1. {G->[]G:0, <>G:0, ~G:0}
I am using a flawed version of Tisthammer’s proof (it best illustrates the tableau method). The initial assumption includes G->G, where -> is material implication (A->B is equivalent to ~A OR B).
The tableau method identifies subsets of this set that can be removed and replaced with other sets of propositions. An atomic proposition or its negation cannot be further reduced, and will remain in the set. Thus, there is nothing to be done about ~G:0.
The rule for possibility is that <>A:i can be removed, and replaced with a pair of propositions {iRj, A:j}. We assume a new world j that is accessible to i via the relation R, and where A is true. The world j “witnesses” the possibility of A at i. Making this change:
2. {G->[]G:0, 0R1, G:1, ~G:0}
Notice that <>G:0 has been removed. The rule for an implication in a world is that the implication is true if either the antecedent is false at the given world or the consequent is true in the given world. This requires that the tableau branches into two possibilities, which must both be explored. The success of either branch leads to a counter-model that makes the inference invalid.
3a. {0R1, G:1, ~G:0}
3b. {[]G:0, 0R1, G:1, ~G:0}
Notice that 3a. actually includes two instances of ~G:0 (one from the initial set, and one from the negation of the antecedent of G->:0). We remove duplicates. First, let’s explore 3b. The rule for necessity is that a proposition A:i is not removed, but propositions A:j are added for every world j such that iRj is already known. A:i is left behind to remind us that we must add A:j everytime we encounter a new j such that iRj. This allows modal tableau to be possibly infinite.
4b. {[]G:0, 0R1, G:1, ~G:0}
results from adding G:1 to the set, due to G:0 and 0R1. Note that G:1 was already included. At this point, there is no rule that we can apply that will add anything new to the set, and there is not yet a contradiction. This means that the inference is not generally true (this set demonstrates a way that the assumptions can be true and the conclusion can be false). However, if we require R to be reflexive, we must add iRi for every known world i.
5b. {[]G:0, 0R1, G:1, ~G:0, 0R0, 1R1}
6b. {[]G:0, 0R1, G:1, ~G:0, 0R0, 1R1, G:0} X
At the last step (6b), we added G:0 from G:0 and 0R0. The X indicates that this branch of the tableau is closed, since it contains a contradiction (~G:0 and G:0) at a world. That means that (along this branch), there is no non-contradictory way that the assumptions can be true and the conclusion false.
Returning ot 3a:
3a. {0R1, G:1, ~G:0}
there is no rule left to apply, and no contradiction. Thus the inference is not valid, and this set provides a counter model: two worlds 0 and 1, with 0R1 and ~G in world 0 and G in world 1.
Variation: Still not Tisthammer’s Argument
What if we replaced G->G with G<->G? The initial set is:
1. {G<->[]G:0, <>G:0, ~G:0}
as before, we eliminate <>G:0
2. {G<->[]G:0, 0R1, G:1, ~G:0}
The rule for A<->B is that either both A and B are true, or they are both false. Thus, the tableau branches again:
3a. {G:0, []G:0, 0R1, G:1, ~G:0} X
3b. {~[]G:0, 0R1, G:1, ~G:0}
Notice that the duplicate ~G:0 was removed from 3b. 3a contains a contradiction. We continue with 3b. All negative modalities are rewritten (~A is <>~A, and ~<>A is ~A):
4b. {<>~G:0, 0R1, G:1, ~G:0}
And the rule for <> is used:
5b. {0R2, ~G:2, 0R1, G:1, ~G:0}
With no rule left be applied and no contradiction. The inference is not valid, and the counter-model can be read from the set 5b.
Tisthammer’s Argument
Finally, let’s consider Tisthammer’s argument:
1. {[](G->[]G):0, <>G:0, ~G:0}
2. {[](G->[]G):0, 0R1, G:1, ~G:0}
we immediately eliminate <>G:0 as before. Next, we use the rule to make G->G at every world known accessible from world 0:
3. {[](G->[]G):0, 0R1, G:1, ~G:0, G->[]G:1}
We break G->G:1 into the two possibilities:
4a. {[](G->[]G):0, 0R1, G:1, ~G:0, ~G:1} X
4b. {[](G->[]G):0, 0R1, G:1, ~G:0, []G:1}
4a contains a contradiction. 4b. does not, and there is no rule that can be applied to add anything new to the set. If we could get G:0 or ~G:1, we would have a contradiction in 4b. There is no immediate way to add ~G:1; but we could derive G:0 from G:1 if 1R0. Let’s add that:
5b. {[](G->[]G):0, 0R1, G:1, ~G:0, []G:1, 1R0}
6b. {[](G->[]G):0, 0R1, G:1, ~G:0, []G:1, 1R0, G:0} X
The way to get 1R0 from set 4b is to insist that the accessibility relation R is symmetric. Since all branches of the tableau are closed (contain a contradiction), there is no way that the premises of the argument could be true and the conclusion could be false in a modal logic with a symmetric accessibility relation.
Conclusion
Thus, the Tisthammer argument is valid (remember that his first axiom is a strict (or necessary) implication). To reject the argument, one must reject at least one of , <>G, and that the accessibility relation on possible worlds is symmetric. I’ve argued in this thread (and I still maintain), that adoption of any two of these nearly compels rejection of the third. I personally find it easiest to reject one of either <>G or symmetry, since I am inclined to grant that Libertarian has made a good case for his .