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Theorem nandsym1 31591
 Description: A symmetry with ⊼. See negsym1 31586 for more information. (Contributed by Anthony Hart, 4-Sep-2011.)
Assertion
Ref Expression
nandsym1 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → (𝜓𝜑))

Proof of Theorem nandsym1
StepHypRef Expression
1 df-nan 1440 . . . . 5 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) ↔ ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
21biimpi 205 . . . 4 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
3 df-nan 1440 . . . . 5 ((𝜓 ⊼ ⊥) ↔ ¬ (𝜓 ∧ ⊥))
43anbi2i 726 . . . 4 ((𝜓 ∧ (𝜓 ⊼ ⊥)) ↔ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
52, 4sylnib 317 . . 3 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
6 simpl 472 . . . 4 ((𝜓𝜑) → 𝜓)
7 fal 1482 . . . . 5 ¬ ⊥
87intnan 951 . . . 4 ¬ (𝜓 ∧ ⊥)
96, 8jctir 559 . . 3 ((𝜓𝜑) → (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
105, 9nsyl 134 . 2 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓𝜑))
11 df-nan 1440 . 2 ((𝜓𝜑) ↔ ¬ (𝜓𝜑))
1210, 11sylibr 223 1 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → (𝜓𝜑))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383   ⊼ wnan 1439  ⊥wfal 1480 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8 This theorem depends on definitions:  df-bi 196  df-an 385  df-nan 1440  df-tru 1478  df-fal 1481 This theorem is referenced by: (None)
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