at least two formally valid arguments are not valid.

Here’s a formal proof of the liar:

1.  (x)(p)((Cx • S_x(p)) ⊃ ~p)

2.  Cc • S_c((x)(Cx • S_x(p) ⊃ ~p))  ∴  ~(x)(p)((Cx • S_x(p)) ⊃ ~p)

3.  Cc • S_c((x)(p)((Cx • S_x(p)) ⊃ ~p) ⊃ ~(x)(p)((Cx • S_x(p)) ⊃ ~p 1, UI

4.  ~(x)(p)((Cx • S_x(p)) ⊃ ~p)   2, 3, MP

In English:

1. for any x and any p if x is a Cretan and x says p then ~p.

2. Chris is a Cretan and Chris says for any x and any p if x is a Cretan and x says p then ~p.

3. If Chris is a Cretan and Chris says for any x and any p if x is a Cretan and x says p then ~p then it’s not the case that for any x and any p if x is a Cretan and x says p then ~p. (From 1 and universal instantiation)

4. it’s not the case that for any x and any p if x is a Cretan and x says p then ~p. (from 2, 3, and modus ponens)

Here might be a way out:

The argument’s form is valid, but there is no time in which the string of symbols which compose lines 1 and 2 express a proposition. Since strings of symbols which fail to express propositions eo ipso fail to express premises, any instance of lines 1 and 2 will fail to constitute an argument.  Since only arguments may be valid and lines 1 and 2 fail to constitute an argument, nothing follows from lines 1 and 2.  Ergo, there is no liar’s paradox.

Objection:  Wow. Way to be ad hoc.

Reply: False. I have independent reasons for saying that some strings of symbols which appear to express propositions don’t actually express propositions. For example, here’s a formal proof of Zeus’ existence:

1. (x)x = x      ∴  (∃x)x = z

2. z = z              1, UI

3. (∃x)x = z      2, EG

In English:

1. Everything is identical to itself.

2. Zeus is Zeus.  (from 1 and universal instantiation)

3. There is something such that it is Zeus.  (from 2, and existential generalization)

I think that this argument’s form is valid, and I also think that its one and only premise is true. However, I don’t think the string of symbols in line 2 expresses a proposition because in order for it to express a proposition, the letter ‘z’ must refer to Zeus. Since Zeus doesn’t exist,  ‘z’ doesn’t name anything, and therefore line 2 is not even false. Since the proof only goes through if each line expresses a proposition, lines 1, 2, and 3 fail to constitute a proof.