1.1 Substitution

• \(f(x) = 2x\)
• \(f(2) = 4\)
• \(f(g(y)) = 4\)
  • \(g = \forall y, y \mapsto 2\)

What does it really mean when we substituted \(2\) for \(x\)?
One way to look at it is we really substituted function \(g(y)\) for \(x\).
function \(g(y)\) maps all numbers to 2. Graphically this is a just a horizontal line on 2.
We can tell substitutions aka \(g\) function is not injective.

1.2 Why do we need to verify solutions?

Keypoint: Some equation simplifications are non-invertible

3 examples of 3 Types of equation reduction:

1. \(eq_1\overset{?}{=}eq_2 \overset{\square^2}{\Longrightarrow} eq_1^2\overset{?}{=}eq_2^2\)
2. \(eq_1\overset{?}{=}eq_2 \overset{\sqrt{\square}}{\Longleftarrow} \sqrt{eq_1}\overset{?}{=}\sqrt{eq_2}\)
3. \(eq_1\overset{?}{=}eq_2 \overset{+2}{\Longleftrightarrow} eq_1+2\overset{?}{=}eq_2+2\)

1.3 Functional identities

Imagine if we had a functional identity \[ \forall a: \forall f : f(a) = 1 \]

The only function that satisfies this is \(f(x) = 1\)

Note \(\{\forall a: \forall f : f(a) = 1\} \neq \{f(x) = 1\}\)

One is a functional identity and the other is a single function.

1.4 Substitution method

Logarithm satisfies \(\forall x,y : \forall f:f(x \cdot y) = f(x) + f(y)\)

• \([assume\ \hat{a}]\)
  • \([let\ \exists \bar{b}, \bar{b}= 1]\)
    • \([assume\ \hat{f}]\)
      • plug in \(\hat{a} , \bar{b}, \hat{f}\) into functional identity
      • \(\hat{f}(\hat{a} \cdot \bar{b}) = \hat{f}(\hat{a}) + \hat{f}(\bar{b})\) , apply substitution \(\bar{b} = 1\)
      • \(\hat{f}(\hat{a} \cdot 1) = \hat{f}(\hat{a}) + \hat{f}(1)\)
      • \(\hat{f}(\hat{a}) = \hat{f}(\hat{a}) + \hat{f}(1)\)
      • conclude \(\hat{f}(1) = 0\)
    • \(\forall f:f(1) = 0\) discharge arbitrary \(\hat{f}\)
  • \(b\) is no longer in context; Simply drop existential quantifier instead of discharging
• \(\forall a : \forall f: f(1) = 0\) discharge arbitrary \(\hat{a}\)

Plug back our conclusion back into original functional identity for verification (not shown here).

Notice \(Log(1) = 0\)

\(f(1) = 0\) is true for any function f that satisfies the identity \(f(x \cdot y) = f(x) + f(y)\)