Theorem nandsym1 36992

Description: A symmetry with ⊼.

See negsym1 36987 for more information. (Contributed by Anthony Hart, 4-Sep-2011.)

Assertion

Ref	Expression
nandsym1	⊢ ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → (𝜓 ⊼ 𝜑))

Proof of Theorem nandsym1

Step	Hyp	Ref	Expression
1		df-nan 1522	. . . . 5 ⊢ ((𝜓 ⊼ (𝜓 ⊼ ⊥)) ↔ ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
2	1	biimpi 219	. . . 4 ⊢ ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ (𝜓 ⊼ ⊥)))
3		df-nan 1522	. . . . 5 ⊢ ((𝜓 ⊼ ⊥) ↔ ¬ (𝜓 ∧ ⊥))
4	3	anbi2i 635	. . . 4 ⊢ ((𝜓 ∧ (𝜓 ⊼ ⊥)) ↔ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
5	2, 4	sylnib 331	. . 3 ⊢ ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
6		simpl 488	. . . 4 ⊢ ((𝜓 ∧ 𝜑) → 𝜓)
7		fal 1584	. . . . 5 ⊢ ¬ ⊥
8	7	intnan 492	. . . 4 ⊢ ¬ (𝜓 ∧ ⊥)
9	6, 8	jctir 530	. . 3 ⊢ ((𝜓 ∧ 𝜑) → (𝜓 ∧ ¬ (𝜓 ∧ ⊥)))
10	5, 9	nsyl 141	. 2 ⊢ ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → ¬ (𝜓 ∧ 𝜑))
11		df-nan 1522	. 2 ⊢ ((𝜓 ⊼ 𝜑) ↔ ¬ (𝜓 ∧ 𝜑))
12	10, 11	sylibr 237	1 ⊢ ((𝜓 ⊼ (𝜓 ⊼ ⊥)) → (𝜓 ⊼ 𝜑))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ⊼ wnan 1521  ⊥wfal 1582

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-nan 1522  df-tru 1573  df-fal 1583

This theorem is used by: (None)