User:Bill stoddart 2/sandbox

∀ x,y,z • (z∈x ⇔ z∈y) ⇒ x=y

Given a function f and


 * $$ \forall x,y \bullet y>x \implies f(y) > f(x) $$

Then we say that f is strictly monotonic.

Math text without indent.

$$\forall A\forall w_1 \forall w_2\ldots \forall w_n \bigl[\forall x ( x\in A \Rightarrow \exists! y\,\phi ) \Rightarrow \exists B \ \forall x \bigl(x\in A \Rightarrow \exists y (y\in B \land \phi)\bigr)\bigr].$$ $$\forall A\forall w_1 \forall w_2\ldots \forall w_n \bigl[\forall x ( x\in A \Rightarrow \exists! y\,\phi ) \Rightarrow \exists B \ \forall x \bigl(x\in A \Rightarrow \exists y (y\in B \land \phi)\bigr)\bigr].$$

Math text with indent


 * $$\forall A\forall w_1 \forall w_2\ldots \forall w_n \bigl[\forall x ( x\in A \Rightarrow \exists! y\,\phi ) \Rightarrow \exists B \ \forall x \bigl(x\in A \Rightarrow \exists y (y\in B \land \phi)\bigr)\bigr].$$
 * $$\forall A\forall w_1 \forall w_2\ldots \forall w_n \bigl[\forall x ( x\in A \Rightarrow \exists! y\,\phi ) \Rightarrow \exists B \ \forall x \bigl(x\in A \Rightarrow \exists y (y\in B \land \phi)\bigr)\bigr].$$
 * $$\forall A\forall w_1 \forall w_2\ldots \forall w_n \bigl[\forall x ( x\in A \Rightarrow \exists! y\,\phi ) \Rightarrow \exists B \ \forall x \bigl(x\in A \Rightarrow \exists y (y\in B \land \phi)\bigr)\bigr].$$

Specifying block mode and how to get the shorter form of implies.

$$ x<y \, \implies \, \Rightarrow f(x) < f(y) $$

Diamond and Nabla $$ S ; T \diamond E \; = \; S \nabla T \nabla E $$