Theorem dissym1 36989

Description: A symmetry with ∨.

See negsym1 36985 for more information. (Contributed by Anthony Hart, 4-Sep-2011.)

Assertion

Ref	Expression
dissym1	⊢ ((𝜓 ∨ (𝜓 ∨ ⊥)) → (𝜓 ∨ 𝜑))

Proof of Theorem dissym1

Step	Hyp	Ref	Expression
1		orc 881	. 2 ⊢ (𝜓 → (𝜓 ∨ 𝜑))
2		falim 1587	. . 3 ⊢ (⊥ → 𝜑)
3	2	orim2i 924	. 2 ⊢ ((𝜓 ∨ ⊥) → (𝜓 ∨ 𝜑))
4	1, 3	jaoi 871	1 ⊢ ((𝜓 ∨ (𝜓 ∨ ⊥)) → (𝜓 ∨ 𝜑))

This proof depends on syntax axioms:   → wi 4   ∨ wo 861  ⊥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-or 862  df-tru 1573  df-fal 1583

This theorem is used by: (None)