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a guest Feb 17th, 2020 71 Never
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  1. Theorem “TP5H20”:
  2.   (s ⇒ u ∧ p) ∧ (s ⇒ w) ∧ ((s ⇒ p) ⇒ t) ∧ (q ∧ r ⇒ s) ∧ (t ⇒ ¬ q) ⇒ (¬(u ∧ p) ∧ q ⇒ ¬ r ∧ ¬(w ⇒ p))
  3. Proof:
  4.   Assuming `s ⇒ u ∧ p`, `s ⇒ w`, `(s ⇒ p) ⇒ t`, `q ∧ r ⇒ s`, `t ⇒ ¬ q`:
  5.     ¬ r ∧ ¬(w ⇒ p)
  6.   ⇐
  7.  
  8.  
  9.     ¬(u ∧ p) ∧ q
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