We suggest a new example of a free subgroup of a group of automaton substitutions. We present two automata with three states which generate this subgroup; the inner semigroups of these automata are not groups.The first example of a free subgroup of the group of automaton substitutions was suggested in [3]; the complete proof of the result was given in [4]. The automata in this example are double-group ones: the inner semigroups of both these automata and the automata inverse to them are groups. Such automata constitute a quite small part of all automata. Moreover, if we consider groups generated by initial subautomata of an automaton, then, as shown in [6], there exist 122 nonisomorphic groups generated by the automata with three states, and only the automaton described in [3] generates a free group. In [5], an automaton torsion-free group (not free) with three generating semigroup automata is given.We construct a free group generated by two semigroup automata. The elements of the group of automaton substitutions AS 2 [1, 2] are one-to-one mappings of a set of words over the alphabet f0; 1g which are realised by initial automata with input and output alphabets f0; 1g. Consider two such automata A and B which have three states each. The diagram of the automaton A is given in Fig. 1; as usual, the circle corresponding to a state contains the state number and the permutation over the input alphabet which is realised in this state. In the initial state 1, the negation N x which is the cycle .0; 1/ is realised, and the identical substitution x is realised in the states 2 and 3.The mapping realised by the automaton A is denoted by f . The inverse mapping f 1 is realised by the automaton N A given in Fig. 2. With the use of standard notation, we write