Have I misunderstood this argument about truth values?

Question

7.2.1 Every truth value is confirmed by intuition that affirms it.

$(∀x)(Tx\to Ix)$

7.2.2. Every intuition that affirms truth value is a feeling of correctness.

$(∀x)(Ix\to Cx)$

7.2.3. The feeling of correctness cannot be proven to be a criterion for determining truth values.

$(∀x)(Cx\to \sim D)$

Ergo, there exists no tool for determining truth values.

$(∀x)\sim D$

I've been given the valid argument above. However, my translation shows an invalid argument, because its conclusion cannot possibly be derived from from its premises. Have I misunderstood the given argument?

Answer 1

Among other things, the argument assumes that determining a truth value, affirming a truth value, and a truth value being confirmed are all materially equivalent. Further the symbolic version of the argument could be more precise; for example,

$\forall x \exists y (Tx \implies (Cx \implies (Iy \land Ayx)))$

could be used instead of the first formula, where ‘C’ means “is confirmed” and ‘A’ means “affirms”.

Answer 2

In the following attempt, the argument has inconsistent premises, so is automatically valid:

• $Px\;{:\Leftrightarrow}\;$ $x$ is a proposition
• $Tx\;{:\Leftrightarrow}\;$ $x$ is a tool
• $Vp\;{:\Leftrightarrow}\;$ proposition $p$ is true
• $Ip\;{:\Leftrightarrow}\;$ intuition says that proposition $p$ is true
• $Fp\;{:\Leftrightarrow}\;$ feeling says that proposition $p$ is true
• $Ltp\;{:\Leftrightarrow}\;$ tool $t$ says that proposition $p$ is true

7.2.1 Every truth value is confirmed by intuition that affirms it.

$\forall p\;\Big(Pp\to(Vp\leftrightarrow Ip)\Big)$

7.2.2. Every intuition that affirms truth value is a feeling of correctness.

$\forall p\;\Big(Pp\to(Ip\leftrightarrow Fp)\Big)$

7.2.3. The feeling of correctness cannot be proven to be a criterion for determining truth values.

$\sim\forall p\;\Big(Pp\to(Fp\leftrightarrow Vp)\Big)$

Ergo, there exists no tool for determining truth values.

Therefore, $\sim\exists t\;\bigg(Tt\land\;\forall p\;\Big(Pp\to(Ltp\leftrightarrow Vp)\Big)\bigg).$