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Theorem nandsym1 34611
Description: A symmetry with .

See negsym1 34606 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 1487 . . . . 5 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) ↔ ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
21biimpi 215 . . . 4 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
3 df-nan 1487 . . . . 5 ((𝜓 ⊼ ⊥) ↔ ¬ (𝜓 ∧ ⊥))
43anbi2i 623 . . . 4 ((𝜓 ∧ (𝜓 ⊼ ⊥)) ↔ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
52, 4sylnib 328 . . 3 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
6 simpl 483 . . . 4 ((𝜓𝜑) → 𝜓)
7 fal 1553 . . . . 5 ¬ ⊥
87intnan 487 . . . 4 ¬ (𝜓 ∧ ⊥)
96, 8jctir 521 . . 3 ((𝜓𝜑) → (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
105, 9nsyl 140 . 2 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓𝜑))
11 df-nan 1487 . 2 ((𝜓𝜑) ↔ ¬ (𝜓𝜑))
1210, 11sylibr 233 1 ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → (𝜓𝜑))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 396  wnan 1486  wfal 1551
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 206  df-an 397  df-nan 1487  df-tru 1542  df-fal 1552
This theorem is referenced by: (None)
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