Second-Order Logic and Fagin`s Theorem
... We must write a first-order formula asserting that Q, S, D encode a correct accepting computation of N . The only difficulty in doing this is that for each move t̄, we must ascertain the symbol ρt̄ that is read by N . ρt̄ is equal to σi where Si (t̄′ ) holds, and t̄′ is the last time before t̄ that ...
... We must write a first-order formula asserting that Q, S, D encode a correct accepting computation of N . The only difficulty in doing this is that for each move t̄, we must ascertain the symbol ρt̄ that is read by N . ρt̄ is equal to σi where Si (t̄′ ) holds, and t̄′ is the last time before t̄ that ...
Adding and Subtracting Fractions
... What if you have to add or subtract fractions with unlike denominators? • Don’t panic. To solve unlike fractions, there are a couple extra steps (like 7 if you want to know…). ...
... What if you have to add or subtract fractions with unlike denominators? • Don’t panic. To solve unlike fractions, there are a couple extra steps (like 7 if you want to know…). ...
AMC 8 Preparation
... Jo can climb maximum 3 stairs and assume that Jo can climb any number of stairs, including all 6 stairs. • Then, we can easily calculate the numbers of combinations of positions of k bars out of total 5 positions (the gaps between 6 stars), ...
... Jo can climb maximum 3 stairs and assume that Jo can climb any number of stairs, including all 6 stairs. • Then, we can easily calculate the numbers of combinations of positions of k bars out of total 5 positions (the gaps between 6 stars), ...
Finite-variable fragments of first
... machine M over a (finite) alphabet A, it is straightforward to construct a tiling system, TM = (C, H, V ), together with a mapping eM : A → C such that, for all N , M halts with output 1 in time at most N on input a0 , . . . , an−1 if and only if TM has a tiling of size N with initial segment eM (a0 ...
... machine M over a (finite) alphabet A, it is straightforward to construct a tiling system, TM = (C, H, V ), together with a mapping eM : A → C such that, for all N , M halts with output 1 in time at most N on input a0 , . . . , an−1 if and only if TM has a tiling of size N with initial segment eM (a0 ...