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Here is the proof that \(^{\omega 2}\omega\) = \(\varepsilon_{\varepsilon_0}\) or a weaker ordinal \(\varepsilon_1\).

Ordinal tetration

Ordinal tetration is a type of normal form but we can express ordinals with 0, 1, \(\omega\), addition, multiplication, exponentiation, and tetration.

Proof

The proof require to have 1 new property, OPTD (\(\omega\) power tower depth). Now what's OPTD? A natural number have OPTD 0, \(\omega\) to maximum value that less than \(\omega^{\omega}\) have OPTD 1, and \(\omega^{\alpha}\) have OPTD \(\beta+1\), such that \(\alpha\) have OPTD \(\beta\).

Also, I also show that \(1+\omega\) = \(\omega\) and \(\omega^{\varepsilon_02}\) have OPTD \(\omega\).

Sub-proof 1 (\(1+\omega\) = \(\omega\))

Imagine 0 is the empty set and \(\alpha+1\) is the union set of all elements of \(\alpha\) and the set of \(\alpha\). Therefore \(\omega\) = \(\{,\{\},\{,\{\}\},\{,\{\},\{,\{\}\}\},...\}\).

However, we start at 1 and increase by 1 each till we reach \(1+\omega\). The result is unexpected and results as the set of \(\omega\). Therefore, I proved that \(1+\omega\) = \(\omega\).

Sub-proof 2 (\(\omega^{\varepsilon_02}\) have OPTD \(\omega\))

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