Sub-Hilbert Lattices

On the variety of strong subresiduated lattices

“…Every filter of A is an implicative filter of A , however the converse is not true in general (cf. [4] for details).…”

“…Let us begin by recalling the systems $$\mathsf {R4}$$ and $$\mathsf {R4}^\star$$ presented by a Hilbert style system in the language $$\mathcal {L} = \lbrace \rightarrow , \vee , \wedge , \lnot \rbrace$$ of type (2, 2, 2, 1), as it was done in [10] and [5] respectively.…”

“…The $$\mathsf {R4}^\star$$ calculus [5] is the calculus, in the language of $$\mathsf {R4}$$, which is defined by the same axiom schemes of $$(\mathsf {R4})$$ by replacing the axiom scheme $$(\mathsf {A2})$$ by the axiom scheme $$(\mathsf {Ax})$$$$(\alpha \rightarrow \beta ) \rightarrow ((\beta \rightarrow \delta ) \rightarrow (\alpha \rightarrow \delta ))$$, and whose rules are modus ponens and the additional rule $$\alpha \vdash \beta \rightarrow \alpha$$, which is called $$(\mathsf {T})$$.…”

“…The axiom scheme $$(\mathsf {A2})$$ of $$\mathsf {R4}$$ is a theorem of $$\mathsf {R4}^\star$$ (cf. [5, Lemma 5.4]), so every deduction in $$\mathsf {R4}$$ is valid in $$\mathsf {R4}^\star$$. On the other hand, in [10] it was mentioned that $$(\mathsf {Ax})$$ is a theorem in $$\mathsf {R4}$$, so every deduction in $$\mathsf {R4}^\star$$ not involving the use of rule $$(\mathsf {T})$$ is valid in $$\mathsf {R4}$$.…”

“…In what follows we will write $$\mathcal {L}$$ for the intuitionistic language $$\mathcal {L} = \lbrace \rightarrow , \vee , \wedge , \lnot \rbrace$$. In [5] it was also shown that $$\mathsf {R4}^\star$$ is an implicative logic in the sense of [11] and that in consequence this logic is complete with respect to the class $$\mathrm{Alg}_{\mathsf {R4}^\star }^\ast$$, where we included the greatest element 1 in the signature of the algebras. Furthermore, for every algebra in $$\mathrm{Alg}_{\mathsf {R4}^\star }^\ast$$ we have that $$0:=\lnot 1$$ is the bottom and the identity $$\lnot x = x \rightarrow 0$$ is satisfied.…”