What does '"turnstile" A-> B' mean?

Question

I apologize for not knowing how to make the acutal symbols.

I believe the turnstile means "proves", so I'm confused what it means when it doesn't have anything before it.

Also, what does '"triangle" union {A}' mean? I believe the triangle means difference between sets, so that expression makes no sense to me.

Is $\vdash$ the turnstile you're talking about? I don't understand what you mean with the other symbol, you can use detexify to find out how to type them — Alessandro Codenotti, Nov 14 '19 at 12:22
Yes, and the point is, what does it mean when there is nothing before it? The triangle is same as delta symbol but for set difference. — csp2018, Nov 14 '19 at 12:32
Yes, but nothing before the triangle. And after the triangle, union with a set. — csp2018, Nov 14 '19 at 12:40
In what context did you encounter that notation? That is, what did the text around that say? — celtschk, Nov 14 '19 at 12:51
(just guessing, since no reference for context was provided) A turnstile with nothing to the left means that no wffs are needed to obtain the wff(s) on the right side (e.g. in ordinary logic we have $;⊢{P ;\vee -P}),$ and probably $\triangle$ denotes a set of wffs. — Dave L. Renfro, Nov 14 '19 at 12:57

score 4 · Answer 1 · answered Nov 14 '19 at 13:06

If $S$ is a set of formulae, then $S \vdash \varphi$ means that $\varphi$ can be derived from the formulae in $S$. Intuitively, $\varphi$ is true under the assumptions in $S$.

$\vdash \varphi$ is shorthand for $\varnothing \vdash \varphi$, meaning that $\varphi$ can be proved without making any assumptions (beyond the axioms of the logical system).

The 'triangle' you speak of is $\Delta$, the capital Greek letter delta, which just stands for an arbitrary set of formulae (like the '$S$' I mentioned above).

So in particular:

$\Delta \cup \{ A \} \vdash B$ means that $B$ can be derived under the assumptions in $\Delta$ and the additional assumption $A$.
$\Delta \vdash A \to B$ means that $A \to B$ can be derived under the assumptions in $\Delta$.
The deduction theorem says that if $\Delta \cup \{ A \} \vdash B$ then $\Delta \vdash A \to B$, i.e. under the assumptions in $\Delta$, if $B$ can be derived from the additional assumption $A$, then $A \to B$ can be derived just from $\Delta$.

The form '$A \vdash B$ implies $\vdash A \to B$' is the special case when $\Delta = \varnothing$. That is, if $B$ can be derived from just the assumption $A$, then $A \to B$ can be derived with no assumptions (beyond the axioms of the logical system).

score 4 · Accepted Answer · answered Nov 14 '19 at 13:06

If nothing precedes the turnstile $\vdash$, what follows is either an axiom or derivable entirely from the axioms. Basically, it means that you don't have to make any assumptions beyond the axioms to derive the following formula. $\vdash\text{(formula)}$ is equivalent to $\emptyset\vdash\text{(formula)}$

The “triangle union” in the Wikipedia article you linked in the comments is not a special notation. The $\Delta$ is just the Greek uppercase letter delta, and here the name of a set, and then $\Delta\cup\{A\}$ has then the obvious interpretation, the union of the set named $\Delta$ with the singleton set $\{A\}$.

Thanks all to great answers, but i chose this for extra effort in the comments — csp2018, Nov 14 '19 at 13:11

What does '"turnstile" A-> B' mean?

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