Extended Rule of Implication

Theorem
Any sequent can be expressed as a theorem.

That is:


 * $$p_1, p_2, p_3, \ldots, p_n \vdash q$$

means the same thing as:


 * $$\vdash p_1 \implies \left({p_2 \implies \left({p_3 \implies \left({\ldots \implies \left({p_n \implies q}\right) \ldots }\right)}\right)}\right)$$

The latter expression is known as the corresponding conditional of the former.

Proof
From the Rule of Conjunction, we note the following.

Any sequent:


 * $$p_1, p_2 \vdash q$$

can be expressed as:


 * $$p_1 \and p_2 \vdash q$$

Also, from the Rule of Simplification, any sequent:


 * $$p_1 \and p_2 \vdash q$$

can be expressed as:


 * $$p_1, p_2 \vdash q$$

Let us take the expression:
 * $$p_1, p_2, p_3, \ldots, p_{n-1}, p_n \vdash q$$

By repeated application of the above, we can arrive at:
 * $$p_1 \and \left({p_2 \and \left({p_3 \and \left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots }\right)}\right)}\right) \vdash q$$

For convenience, we can use a substitution instance to substitute $$r_1$$ for:
 * $$p_2 \and \left({p_3 \and \left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots}\right)}\right)$$

to get:
 * $$p_1 \and r_1 \vdash q$$

From the Rule of Implication, we get:
 * $$\vdash \left({p_1 \and r_1}\right) \implies q$$

Using the Rule of Exportation, we then get:
 * $$\vdash p_1 \implies \left({r_1 \implies q}\right)$$

Then we can substitute back for $$r_1$$, to get:
 * $$\vdash p_1 \implies \left({\left({p_2 \and \left({p_3 \and \left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots}\right)}\right)}\right) \implies q}\right)$$

Now we make a substitution of convenience again. We substitute $$r_2$$ for:
 * $$p_3 \and \left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots}\right)$$

We can then take the expression:
 * $$\left({p_2 \and r_2}\right) \implies q$$

and express it as:
 * $$p_2 \and r_2 \vdash q$$

and use the Rule of Exportation to get:
 * $$p_2 \implies \left({r_2 \implies q}\right)$$

which, substituting back for $$r_2$$, gives us:
 * $$p_2 \implies \left({\left({p_3 \and \left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots}\right)} \right) \implies q}\right)$$

Similarly:
 * $$\left({p_3 \and \left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots}\right)}\right) \implies q$$

converts to:
 * $$p_3 \implies \left({\left({\ldots \and \left({p_{n-1} \and p_n}\right) \ldots}\right) \implies q}\right)$$

The pattern becomes apparent. Eventually we reach:
 * $$\left({p_{n-1} \and p_n}\right) \implies q$$

which converts to:
 * $$p_{n-1} \implies \left({p_n \implies q}\right)$$

We can substitute these back into our original expression and get:
 * $$\vdash p_1 \implies \left({p_2 \implies \left({p_3 \implies \left({\ldots \implies \left({p_n \implies q}\right) \ldots }\right)}\right)}\right)$$

Comment
This shows us that every sequent containing the symbol $$\vdash$$ can, if so desired, be expressed in the form of a theorem which has $$\implies$$.

Some authors ignore the $$\vdash$$ concept, and construct their exposition of PropLog using $$\implies$$ instead. This is a completely valid approach, and appears to have been gaining favour in more recent years, particularly in the field of computability.