Extended Rule of Implication

Context
Natural deduction

Definition
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:

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

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

Proof
First we note the following.

Any sequent:

$$p_1, p_2 \vdash q$$

can be expressed as:

$$p_1 \land p_2 \vdash q$$

This follows directly from the Rule of Conjunction.

Also, any sequent:

$$p_1 \land p_2 \vdash q$$

can be expressed as:

$$p_1, p_2 \vdash q$$

This follows directly from the Rule of Simplification.

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:

$$p_1 \land \left({p_2 \land \left({p_3 \land \left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots }\right)}\right)}\right) \vdash q$$

For convenience, we can use a substitution instance to substitute $$r_1$$ for:

$$p_2 \land \left({p_3 \land \left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots}\right)}\right)$$

to get:

$$p_1 \land r_1 \vdash q$$

From the Rule of Implication, we get:

$$\vdash \left({p_1 \land r_1}\right) \Longrightarrow q$$

Using the Rule of Exportation, we then get:

$$\vdash p_1 \Longrightarrow \left({r_1 \Longrightarrow q}\right)$$

Then we can substitute back for $$r_1$$, to get:

$$\vdash p_1 \Longrightarrow \left({\left({p_2 \land \left({p_3 \land \left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots}\right)}\right)}\right) \Longrightarrow q}\right)$$

Now we make a substitution of convenience again. We substitute $$r_2$$ for:

$$p_3 \land \left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots}\right)$$

We can then take the expression:

$$\left({p_2 \land r_2}\right) \Longrightarrow q$$

and express it as:

$$p_2 \land r_2 \vdash q$$

and use the Rule of Exportation to get:

$$p_2 \Longrightarrow \left({r_2 \Longrightarrow q}\right)$$

which, substituting back for $$r_2$$, gives us:

$$p_2 \Longrightarrow \left({\left({p_3 \land \left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots}\right)} \right)\Longrightarrow q}\right)$$

Similarly:

$$\left({p_3 \land \left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots}\right)}\right) \Longrightarrow q$$

converts to:

$$p_3 \Longrightarrow \left({\left({\ldots \land \left({p_{n-1} \land p_n}\right) \ldots}\right) \Longrightarrow q}\right)$$

The pattern becomes apparent. Eventually we reach:

$$\left({p_{n-1} \land p_n}\right) \Longrightarrow q$$

which converts to:

$$p_{n-1} \Longrightarrow \left({p_n \Longrightarrow q}\right)$$

We can substitute these back into our original expression and get:

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

Q.E.D.

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 $$\Longrightarrow$$.

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