## Philosophy Lexicon of Arguments | |||

Author | Item | Excerpt | Meta data |
---|---|---|---|

Hintikka, J. Books on Amazon |
Game-theoretical Semantics | I 25 Definition game-theoretical semantics/GTS/notation/Hintikka: here the truth of a proposition S in a model M is explained as the existence of a winning strategy in a game of verification (semantic game G (S)). I: I am the verifier. Nature/Opponent: nature or opponent is the falsifier. --- I 26 Rules: (G.E) when the game of the sentence (Ex) has reached S[x] and M, I choose an individual, e.g. b from the domain do/(M) of W. Then the game is continued with respect to S 8b] and M. (G.U.) in the same way, except that here nature chooses the individual b. (G.v) G(S1 v S2) (played in M) starts with my choice of vo Si (i = 1 or 2) The rest of the game is G (Si) (played in the same model m) G.&) Similarly, except that the nature selects Si. (G.~) G (~S) is like G (S) except that the rules of the two players (I and nature) have been exchanged. (G.K.) When the game reaches the set {b} KS and the model (possible world) M0, nature chooses an epistemic b-alternative M1 to M0. The game continues with regard to S and M1. Game-theoretical semantics/GTS/Hintikka: with game-theoretical semantics the semantics can be executed explicitly for branched formulas like (4.6.). They show the informational independence. In (4.6), the steps associated with "(Ex) and ([b] K" are made without the knowledge of the other step. In general, each step is linked to an information set that contains the other steps the player knows when he takes the step. Order: therefore, the structure of the operators of a sentence does not always have to be ordered partially at all. ((s). That is, the order of (Ex) and "knows" may be arbitrary. --- I 27 Game-theoretical semantics/Informational Independence/Hintikka: game-theoretical semantics shows how other basic concepts of a language can also be independent of epistemic operators. For example, an atomic predicate A (x) or a name can be judged independently in M of an epistemic operator, e.g. "knows". ((s) "b knows that x runs" (but not that it is Paul, though x = Paul)). Solution/Hintikka: since the actual referents must be assigned in the winning strategy, expressions like A (x) / {b} K) and A / {b} K actually pick out the actual referents in M0. |
Hin I Jaakko and Merrill B. Hintikka The Logic of Epistemology and the Epistemology of Logic Dordrecht 1989 W I J. Hintikka/M. B. Hintikka Untersuchungen zu Wittgenstein Frankfurt 1996 |

> Counter arguments against **Hintikka**

> Counter arguments in relation to **Game-theoretical Semantics**

back to list view | > Suggest your own contribution | > Suggest a correction

Ed. Martin Schulz, access date 2017-03-27