EQ 2: Textual Replacement

prev: Table of Precedences


E [x := R] or $E _{R} ^{x}$ denotes an expression that is the same as E, but with all the occurrences of x replaced by "(R)". Unnecessary parentheses maybe removed after substitution.

e.g: (z + y) [z := 5] becomes ((5) + y) becomes (by removing unnecessary parentheses) (5 + y)

If x is a list of *distinct* $x_1, x_2,...x_n$ of variables and R a list $R_1\ , R_2 , ... R_n$ of expressions then $E _{R} ^{x}$ denotes the *simultaneous* replacement in E of the variables of x by corresponding expressions of R, each expression being enclosed by parentheses.

(z + y) [z,y := 5,6] becomes ((5) + (6)) becomes (by removing unnecessary parentheses) (5 + 6)

next: Inference Rules

EQ 1: Precedences

Precedences (from high to low)

(1) [x := e] (Textual Substitution)

(2) . (function application)

(3) Unary Prefix Operators + - $\neg$ # ~ $\mathcal { P }$

(4) **

(5) $\ast$ / $\div$ mod gcd

(6) + - $\cup$ $\cap$ $\times$ $\circ$ $\bullet$

(7) $\downarrow$ $\uparrow$

(8) #

(9) $\triangleright$ $\triangleleft$ ^

(10) = < > $\epsilon$ $\subset$ $\subseteq$ $\supset$ $\supseteq$ |

(11) $\vee$ $\wedge$

(12) $\Rightarrow$ $\Leftarrow$

(13) $\equiv$

Notes: All non associative prefix binary operators associate to the left, except ** , $\triangleleft$ and $\Rightarrow$ which associate to the right.


All operators on lines 10, 12 and 13 may have a slash / through them to denote negation.

Thus b $\not\equiv$ c is equivalent to $\neg$ ( b $\equiv$ c )


A core subset of the above, omitting unnecessary symbols, which is used to prove the theorems of the logic itself is as follows

(a) [x := e] (Textual Substitution)

(b) . (function application)

(c) $\neg$

(d) $\ast$ /

(e) =

(f) $\vee$ $\wedge$

(g) $\Rightarrow$ $\Leftarrow$

(h) $\equiv$

prev: next: Textual Substitution

Testing Latex Rendering

Just some random text


$$1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for} |q|<1.$$


$$\begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}$$


embedded latex

More MathJax tests. Embedding formulae inside sentences

The formula $\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for} |q|<1.$ embedded in the middle of a sentence

This expression $\sqrt{3x-1}+(1+x)^2$ is an example of an inline equation.



The question is Prove Universal Instantiation i.e

Prove $(\forall x |: P) \Rightarrow P [ x := E ]$

LHS $\hspace{2 cm } (\forall x |: P)$

= $\hspace{0.5 cm }$ notation expansion

$\hspace{2 cm } (\forall x | true : P)$

= $\hspace{0.5 cm }$ true $\equiv a \vee \neg a$

$\hspace{2 cm } (\forall x | (x = E) \vee (x \neq E) : P)$

= $\hspace{0.5 cm }$ Axiom: Range Split $(\star | R \vee S: P) \equiv (\star | R : P) \star (\star | S : P)$ here $\star = \wedge.$ notated as $\forall$


$\hspace{2 cm } (\forall x | (x = E) : P) \wedge (\forall x | (x \neq E) : P)$

$\Rightarrow \hspace{0.5 cm } Theorem: a \wedge b \Rightarrow a$

$\hspace{2 cm } (\forall x | (x = E) : P)$

= $\hspace{0.5 cm }$ Single Point Axiom $(\star y | (y = F) : Q) \equiv$ Q [ y := F ]

$\hspace{2 cm }$ P [ x := E ] = RHS.
QED