Using the principles of strong induction (thank you zacin bas$d God) We can consider the initial case, where step has 3 letters. 1-S 2-T 3-E 4-P .......... Oh my gawd, step 4 confirmed
Using the principles of strong induction (thank you zacin bas$d God) We can consider the initial case, where step has 3 letters. 1-S 2-T 3-E 4-P .......... Oh my gawd, step 4 confirmed
Case 1: STEP is a subset of Stephen Siklos. Case 2: STEP I, II, III are elements of the subset STEP.
Since STEP I, II, III are subsets of STEP, and due to the fact that STEP is a subset of Stephen Siklos, then there must be a distinct subset of elements of the set Stephen Siklos that are members of it. Therefore, using proof by exhaustion, the proposition that Stephen Siklos is hiding STEP IV is true. QED
I'm so excited to be the first half of the day before I get a follow back on my way home from work (spammed suggestive text and couldn't stop because sentence was interesting)
I'm so excited to be the first half of the day before I get a follow back on my way home from work (spammed suggestive text and couldn't stop because sentence was interesting)
Case 1: STEP is a subset of Stephen Siklos. Case 2: STEP I, II, III are elements of the subset STEP.
Since STEP I, II, III are subsets of STEP, and due to the fact that STEP is a subset of Stephen Siklos, then there must be a distinct subset of elements of the set Stephen Siklos that are members of it. Therefore, using proof by exhaustion, the proposition that Stephen Siklos is hiding STEP IV is true. QED
Gonna take a page from my boy ramanujan and do a proof by intuition: Consider how cool primes are(very) Consider that they're are 3 step papers, a prime, it follows that Siklos would only work in prime number of step papers so there MUST be a FIFTH step paper too
Gonna take a page from my boy ramanujan and do a proof by intuition: Consider how cool primes are(very) Consider that they're are 3 step papers, a prime, it follows that Siklos would only work in prime number of step papers so there MUST be a FIFTH step paper too
Axiom 1: Stephen Siklos is the set with subset of STEP. Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.
By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED
Axiom 1: Stephen Siklos is the set with subset of STEP. Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.
By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED
1. STEP is made up of three letters. 2. STEP is also made up of three papers - I, II, III.
However, STEP is not made up of three letters, so by using proof by contradiction it has been proven that STEP is also not made up of three papers. STEP IV is being secretly hidden by Siklos confirmed. QED Illuminati controls STEP confirmed.
Axiom 1: Stephen Siklos is the set with subset of STEP. Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.
By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED
There are a bunch of STEP papers, possibly uncountably many, Each step paper has a symmetrical domain about the point q=7, this each step paper can be formed by an odd and even function (but **** functions let's use numbers) We know this is true for any step n where n is greater than or equal to one, by induction which I'm too lazy to type out