Proposition 4 (Lack of distance decrement implies vanishing)If are -valued random variables, with the property that for all -valued random variables and some sufficiently small absolute constant , then one can derive a contradiction. Indeed, we may assume from the above proposition that ...
In the first interview session, the participant began by completing a practice problem to become accustomed to the interview format. The participant was then given one of the study tasks. The participant was permitted to work on a proof until he or she wrote a proof that he or she was...
Mathematics of Control Signals & SystemsRepresentation of shift-invariant operators on L2 by H∞ transfer functions: an elementary proof, a generalization to Lp, and a counterexample for L - Weiss - 1991 () Citation Context ...er. It is well-known that in the Hilbert space setting input-...
13In fact, if there is no counterexample to a statement, the statement is valid: this way of thinking can be in its turn an example of backward reasoning. 240 F. Arzarello and C. Soldano generally acts in the following way: he produces a configuration which, provided there is the ...
Such sort-increasing rules could result in a lack of completeness of the computation mechanism, and this is why it may seem wise to add the inductive property stating that x*Conj (x) has sort Nat. This is actually not necessary, as discussed later. Figure 2 shows how a bounded stack of...
13In fact, if there is no counterexample to a statement, the statement is valid: this way of thinking can be in its turn an example of backward reasoning. 240 F. Arzarello and C. Soldano generally acts in the following way: he produces a configuration which, provided there is the ...
: The importance of boundary cases between example and counterexample. In Proceedings of the 35th Conference of the International Group for the Psychology of Mathematics Education, Ankara, Turkey, 10–15 July 2011; Ubuz, B., Ed.; PME: Ankara, Turkey, 2011; Volume 3, pp. 89–96. [Google...
The counterexample in the form of the 𝑑1d1 process disproves the implication 𝑍1∣→𝑍2Z1∣→Z2, despite pointing to the 𝑑′d′ process that would allegedly prove this implication by contradiction. So, in order to prove the 𝑍1∣→𝑍2Z1∣→Z2 implication of the principles...