symbolic contingency

□∃q→∃u :
1. ∀p∃(q→p).

2. ∄u , ∄u→∃u.

c.
∃u→∃(q→u).
∃(∁u→u) ⇔ ∃(∁(∃⊻∄)→u) ⇔ q.
∁u⇔¬∄q.
□∃q→∃u.


∀p∃(q→p) ⇔ ( ∀p∃(q→p) ∧ ¬(p□∃) ) :
1. ∀p∃(q→p)

2. p□∃ ⇔ ( ¬∃(q→p) → q∧¬q ).

c.
∃(q→p) → ¬(p□∃).
∀p∃(q→p) ⇔ ( ∀p∃(q→p) ∧ ¬(p□∃) )




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1. "∀p∃(q→p). " [The Principle of Sufficient Reason & Causal Principle]
2. "∄u , ∄u→∃u" [the set of emergent existents began to exist]
c. "∃u→∃(q→u). [1, 2, Modus Ponens.]
∃(∁u→u) ⇔ ∃(∁(∃⊻∄)→u) ⇔ q. [Extrinsic specifier (q), outside of the set of emergent existents, caused the emergence of the set of emergent existents.]
∁u⇔¬∄q. [The extrinsic specifier exists, not in the set of existents that accepts negation of their existence.]
□∃q. [The extrinsic specifier always exists; it exists necessarily.]"


Let me clarify the corrected steps:

1. The Principle of Sufficient Reason & Causal Principle: "∀p∃(q→p)."
2. The set of emergent existents began to exist: "∄u , ∄u→∃u."
3. Using Modus Ponens on steps 1 and 2: "∃u→∃(q→u)."
4. The extrinsic specifier (q), outside of the set of emergent existents, caused the emergence of the set of emergent existents: "∃(∁u→u) ⇔ ∃(∁(∃⊻∄)→u) ⇔ q."
5. The extrinsic specifier exists, not in the set of existents that accepts negation of their existence: "∁u⇔¬∄q."
6. The extrinsic specifier always exists; it exists necessarily: "□∃q."



1. ∀p∃(q→p)
This is the given premise. It states that for every proposition p, there exists a proposition q such that q implies p.
2. p□∃ ⇔ ( ¬∃(q→p) → q∧¬q ).
This is the contrapositive form of the law of excluded middle, which states that for any proposition p, either p or ¬p is true, and cannot both be false. If p always exists (p □∃), then for any proposition q, we have both q and ¬q, because p can be either true or false.
3. ∃(q→p) → ¬(p□∃).
This follows from the contrapositive of 2, since if there exists a q such that q implies p, then it must not be the case that p always exists.
4. Therefore,
∀p∃(q→p) ⇔ ( ∀p∃(q→p) ∧ ¬(p□∃) )
This follows from 1 and 3, since if for every p, there exists a q such that q implies p (1), then it must not be the case that p always exists (3), which is equivalent to the right-hand side of the equivalence.