It has been shown that, for all rational numbers r such that 0≤ r ≤ 1, the ℵ0-valued Łukasiewicz propositional calculus whose designated truth-values are those truth-values x such that r ≤ x ≤ 1 may be formalised completely by means of finitely many axiom schemes and primitive rules of procedure. We shall consider now the case where r is rational, 0≥r≤1 and the designated truth-values are those truth-values x such that r≤x≤1.
We note that, in the subcase of the previous case where r = 1, a complete formalisation is given by the following four axiom schemes together with the rule of modus ponens (with respect to C),
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU1.gif?pub-status=live)
the functor A being defined in the usual way. The functors B, K, L will also be considered to be defined in the usual way. Let us consider now the functor Dαβ such that if P, Dαβ take the truth-values x, dαβ(x) respectively, α, β are relatively prime integers and r = α/β then
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU2.gif?pub-status=live)
It follows at once from a theorem of McNaughton that the functor Dαβ is definable in terms of C and N in an effective way. If r = 0 we make the definition
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU3.gif?pub-status=live)
We note first that if x ≤ α/β then dαβ(x)≤(β + 1)α/β − α = α/β. Hence
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqn1.gif?pub-status=live)
Let us now define the functions dnαβ(x) (n = 0,1,…) by
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU4.gif?pub-status=live)
Since
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU5.gif?pub-status=live)
it follows easily that
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU6.gif?pub-status=live)
and that
![](//static.cambridge.org/content/id/urn%3Acambridge.org%3Aid%3Aarticle%3AS0022481200049549/resource/name/S0022481200049549_eqnU7.gif?pub-status=live)
Thus, if x is designated, x − α/β > 0 and, if n > − log(x − α/β)/log(β + 1), then (β + 1)n(x−α/β) > 1.