The theorem mp actually has 4 arguments. The first two are the
substitutions, which are usually suppressed. That is, you first have to
prove wff P, then wff Q, then |- P, then |- ( P -> Q ), and then theorem mp
applies to derive |- Q. If you use metamath.exe, "show proof th1" will list
only 5 steps, but with suspicious gaps in the numbering; "show proof th1
/full" will show all the substitution steps (which in fact make up the
majority of the actual proof on disk).

A proof of "wff P" is equivalent to a proof that P is a well formed
formula, and it is this that prevents invalid substitutions like Q := `t =
t )`; you will not be able to prove `wff t = t )` so that proof will not
work. But in any case the verifier isn't being asked to come up with the
substitution (although most metamath proof assistants will, using
unification), so no matching algorithm is necessary, only a substitution
and an equality check.

Mario

On Wed, Oct 7, 2020 at 4:58 PM 'Alexandre Frechette' via Metamath <
[email protected]> wrote:

> I am not sure I understand how metamath guarantees a unique substitution
> at all times. For example, in the proof of `th1` in `demo0.mm`, the two
> following expressions need to be matched in step 32:
>
> `|- ( P -> Q )`
> `|- ( ( t + 0 ) = t -> ( ( t + 0 ) = t -> t = t ) )`
>
> This can be done with two naive substitutions:
>
>  P := `( t + 0 ) = t`
>  Q := `( ( t + 0 ) = t -> t = t )`
>
>  or
>
>  P := `( t + 0 ) = t -> ( ( t + 0 ) = t`
>  Q := `t = t )`
>
> Is there a rule / principle in Metamath I am missing that precludes the
> second ill-defined substitution?
>
> Thank you!
>
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