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Theorem nandsym1 32929
Description: A symmetry with .

See negsym1 32924 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 1610 . . . . 5 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) ↔ ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
21biimpi 208 . . . 4 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
3 df-nan 1610 . . . . 5 ((𝜓 ⊼ ⊥) ↔ ¬ (𝜓 ∧ ⊥))
43anbi2i 617 . . . 4 ((𝜓 ∧ (𝜓 ⊼ ⊥)) ↔ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
52, 4sylnib 320 . . 3 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
6 simpl 475 . . . 4 ((𝜓𝜑) → 𝜓)
7 fal 1668 . . . . 5 ¬ ⊥
87intnan 481 . . . 4 ¬ (𝜓 ∧ ⊥)
96, 8jctir 517 . . 3 ((𝜓𝜑) → (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
105, 9nsyl 138 . 2 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓𝜑))
11 df-nan 1610 . 2 ((𝜓𝜑) ↔ ¬ (𝜓𝜑))
1210, 11sylibr 226 1 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → (𝜓𝜑))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 385  wnan 1609  wfal 1666
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 199  df-an 386  df-nan 1610  df-tru 1657  df-fal 1667
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator