Determinism as Flux
Posted: Mon Apr 10, 2017 8:59 pm
Determinism as Flux
The argument presented will be that strict determinism, due to an inherent element of flux through reflectivity, cannot be observed therefore a probabilistic determinism is the most likely observation of causality. The problem will be observed from the analytic method of philosophy.
1) All causes (Φ) manifest further causes:
(Φα → Φ1 → Φ2 → Φ3)
2) All causes are a beginning cause for further causes:
(Φ1α → Φ1a → Φ1b → Φ1c)
(Φ2α → Φ2a → Φ2b → Φ2c)
etc.
3) Causes manifesting further causes are equal in definition (≜) to causes reflecting (≡) causes, because all manifestation is an element of (∈) reflection:
(Φα → Φ1 → Φ2 → Φ3) ≜ (Φα ≡ Φ1 ≡ Φ2 ≡ Φ3)
∵ (Φx → Φy) ∈ (Φx ≡ Φy)
4) Therefore all beginning causes are self-reflective causes:
(Φα) → (Φα ≡ Φα) → (Φα → Φ1)
5) Reflective causes are equal in definition to similar (~) causes:
(Φα ≡ Φ1 ≡ Φ2 ≡ Φ3) ≜ (Φα ~ Φ1 ~ Φ2 ~ Φ3)
6) Similar causes implies (⊃) a difference in a degree of structure; therefore non-equality:
(Φα ~ Φ1 ~ Φ2 ~ Φ3) ⊃ (Φα ≠ Φ1 ≠ Φ2 ≠ Φ3)
∵ Φd(1,2,3)
7) Reflection manifests a non-equality in definition; therefore all reflection manifests flux:
(Φα ≡ Φ1 ≡ Φ2 ≡ Φ3) → (Φα ≠ Φ1 ≠ Φ2 ≠ Φ3)
∴ ∀(Φ ≡ Φx) → ΔΦ
8 ) All causality is equal in definition to self-propagating flux (Δ):
∀(Φ → Φx) ≜ ΔΦ
9) All causal flux implies non-proportionality (-∝) between causes therefore a deficiency in causal structure (Ω):
∀(ΔΦ) ⊃ -∝(Φ,Φx)
∴ -ΦΩ
10) All deficiencies in observable (O) structure are equal to randomness (ξ); therefore causal structures contain an element of (∋) randomness as deficiency:
∀(-OΩ) = ξ
∴ ∀(ΦΩ) ∋ ξ
11) However all causal structures are equal in definition to stability (◻) as existence:
∀(ΦΩ) ≜ ◻
12) So a contradiction (↯) in the nature of causality occurs unless viewed from a perspective of duality:
(∀(ΦΩ) ≜ ◻) ↯ (∀(ΦΩ) ∋ ξ)
∀(ΦΩ) ⊃ Φ⟨◻|ξ⟩
13) A synthesis (∪) of these dualistic elements within all causal structure manifests as probabilism; therefore all deterministic systems are probabilistic (ϕ):
(Φ◻) ∪ (Φξ) → Φϕ
True?
False?
Maybe?
The argument presented will be that strict determinism, due to an inherent element of flux through reflectivity, cannot be observed therefore a probabilistic determinism is the most likely observation of causality. The problem will be observed from the analytic method of philosophy.
1) All causes (Φ) manifest further causes:
(Φα → Φ1 → Φ2 → Φ3)
2) All causes are a beginning cause for further causes:
(Φ1α → Φ1a → Φ1b → Φ1c)
(Φ2α → Φ2a → Φ2b → Φ2c)
etc.
3) Causes manifesting further causes are equal in definition (≜) to causes reflecting (≡) causes, because all manifestation is an element of (∈) reflection:
(Φα → Φ1 → Φ2 → Φ3) ≜ (Φα ≡ Φ1 ≡ Φ2 ≡ Φ3)
∵ (Φx → Φy) ∈ (Φx ≡ Φy)
4) Therefore all beginning causes are self-reflective causes:
(Φα) → (Φα ≡ Φα) → (Φα → Φ1)
5) Reflective causes are equal in definition to similar (~) causes:
(Φα ≡ Φ1 ≡ Φ2 ≡ Φ3) ≜ (Φα ~ Φ1 ~ Φ2 ~ Φ3)
6) Similar causes implies (⊃) a difference in a degree of structure; therefore non-equality:
(Φα ~ Φ1 ~ Φ2 ~ Φ3) ⊃ (Φα ≠ Φ1 ≠ Φ2 ≠ Φ3)
∵ Φd(1,2,3)
7) Reflection manifests a non-equality in definition; therefore all reflection manifests flux:
(Φα ≡ Φ1 ≡ Φ2 ≡ Φ3) → (Φα ≠ Φ1 ≠ Φ2 ≠ Φ3)
∴ ∀(Φ ≡ Φx) → ΔΦ
8 ) All causality is equal in definition to self-propagating flux (Δ):
∀(Φ → Φx) ≜ ΔΦ
9) All causal flux implies non-proportionality (-∝) between causes therefore a deficiency in causal structure (Ω):
∀(ΔΦ) ⊃ -∝(Φ,Φx)
∴ -ΦΩ
10) All deficiencies in observable (O) structure are equal to randomness (ξ); therefore causal structures contain an element of (∋) randomness as deficiency:
∀(-OΩ) = ξ
∴ ∀(ΦΩ) ∋ ξ
11) However all causal structures are equal in definition to stability (◻) as existence:
∀(ΦΩ) ≜ ◻
12) So a contradiction (↯) in the nature of causality occurs unless viewed from a perspective of duality:
(∀(ΦΩ) ≜ ◻) ↯ (∀(ΦΩ) ∋ ξ)
∀(ΦΩ) ⊃ Φ⟨◻|ξ⟩
13) A synthesis (∪) of these dualistic elements within all causal structure manifests as probabilism; therefore all deterministic systems are probabilistic (ϕ):
(Φ◻) ∪ (Φξ) → Φϕ
True?
False?
Maybe?