Consider the following implicational formulae: A1 = (p  p);A2 = (((p  p)  p)  p) = ((A1  p)  p);


; Ai+1 = ((Ai  p)  p);


; (i = 1; 2; 3;


) Using these formulae (axioms), we may construct the following logics: L1 =< A2i >;L2 =< (A2i  A2i+1) >; L3 =< A2i-1 >;L4 =< (A2i-1  A2i) >; i = 1; 2; 3;


viz. the logic L1 is generated by the axioms A2i, i = 1; 2; 3; :::; the process is analogous for logics L2;L3;L4. The rule of deduction for these logics is unique - modus ponens: A; (A  B) ` B (if the formulae A and (A  B) 2 to the given logic, then formula B also 2 to this logic). Let S be the lattice generated by the logics L1;L2; :L3;L4. The following results are obtained: 1. Lattice S is in nite. 2. If logics L1;L2; :L3;L4 possess a nite number of axioms (viz. i = 1, 2, 3, ..., n), then the respective lattice S is nite. 3. For lattice S the problem of the equality of any two lattice elements is solvable. 4. If the rule of deduction - the substitution is added to the above logics, then statements 1){3) are also true. (The rule of deduction the substitution means: if formula A 2 to the given logic, then the result of the substitution in formula A of any implicational formula of the variable p for the same variable p also 2 to the same logic.)