Teachers in the middle ages supposedly tested the realtime propositional logic a

Question

Teachers in the middle ages supposedly tested the realtime propositional logic ability of a student using an obligato game. In an obligato game, a number of rounds is set and in each round the teacher gives the student successive assertions that the student must either accept or reject as they are given. When the student accepts an assertion, it is added as a commitment; when the student rejects an assertion, its negation is added as a commitment. The student passes the test if the consistency of all commitments is maintained throught the test.

1. Suppose that in a 3-round obligato game, the teacher gives the student the propositions p q r, followed by ¬(p = q) r, and finally (r p) = ¬q. For which of the eight possible sequences of three answers will the student pass the test?

2. Explain why every obligato game has a winning strategy.

Explanation / Answer

1. In the given statement Let p be the proposition "I will do every exercise in this book" and q be the proposition "I will get an "A" in this course." Step 2: We have to express each of these as a combination of p and q. Logical symbol connectives which we are using to express the statement : Negative : ¬ Implication: Biconditional : Conjunction: disjunction: a) I will get an "A" in this course only if I do every exercise in this book. Explanation :if p and q are proposition, then the pq is conditional statement “ if p,then q” Therefore , In this statement is the inverse of p q So we can express ¬p ¬q b) I will get an "A" in this course and I will do every exercise in this book. Explanation: This statement related to the conjunction. If q is true and p is also true, then the conjunction of p and q is also true So we can express q ^ p c) Either I will not get an "A" in this course or I will not do every exercise in this book. Explanation: In this case we can see that the statement of p and q both statement are negative, so we can express ¬q ¬q or ¬q ¬q d) For me to get an "A" in this course it is necessary and sufficient that I do every exercise in this book. Explanation : This is biconditional condition because both p and q are related to each other then the proposition are true. p q p q can also be considered for (P if and only if q ) or ( If p then q or if q then p ).

2.

At the beginning of the game, the student has no commitments, which is trivially consistent.

Then suppose at a given point of the game, the commitments made up to that point are consistent. If accepting the next assertion yields a consistent set of commitments, then the student should accept. If accepting the next assertion made an inconsistent set of commitments, then the negation of that assertion would be provable, given the previous set of commitments. In this case, adding the negation of the assertion will preserve consistency (since nothing new would be able to be proven).

In the above, we used two rules. The first was actually the law of the excluded middle:
"Every set of commitments is either consistent or inconsistent"
and the second was the definition of the negation.

Note that a set of propositions is consistent if and only if there does not exist a proposition whose assertion and negation are both provable from the set of propositions. It should be easy enough to show that there does not exist a proposition whose assertion and negation are both provable from a finite set of propositions if and only if the conjunction of all finitely many propositions is not a contradiction.

Consider the first arbitrary proposition. Since the empty set is consistent (no contradiction is provable from it), a contradiction can only arise if the proposition itself is one. If it is a contradiction, reject it; if it is a tautology, accept it; otherwise, assign it whatever value you wish since it will not on its own be a contradiction.

Consider the second arbitrary proposition. Since the set of commitments contains only one commitment and it is not a contradiction, a contradiction can only arise if the conjunction of the commitment and this arbitrary proposition is a contradiction. If the conjunction of the first commitment and this proposition is a contradiction, reject the proposition; else, accept it.

Let your inductive hypothesis be that after n rounds of Obligato, your commitments form a consistent set (of cardinality less than or equal to n, and ergo finite). Can you demonstrate that it is yet possible to keep the set consistent after the addition of an (n + 1)th round, and consequently keep any previously-consistent set of commitments consistent after any number of rounds?

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