Theorem biluk 899

Description: Lukasiewicz's shortest axiom for equivalential calculus. Storrs McCall, ed., Polish Logic 1920-1939 (Oxford, 1967), p. 96. (Contributed by NM, 10-Jan-2005.)

Assertion

Ref	Expression
biluk	⊢ ((φ ↔ ψ) ↔ ((χ ↔ ψ) ↔ (φ ↔ χ)))

Proof of Theorem biluk

Step	Hyp	Ref	Expression
1		bicom 191	. . . . 5 ⊢ ((φ ↔ ψ) ↔ (ψ ↔ φ))
2	1	bibi1i 305	. . . 4 ⊢ (((φ ↔ ψ) ↔ χ) ↔ ((ψ ↔ φ) ↔ χ))
3		biass 348	. . . 4 ⊢ (((ψ ↔ φ) ↔ χ) ↔ (ψ ↔ (φ ↔ χ)))
4	2, 3	bitri 240	. . 3 ⊢ (((φ ↔ ψ) ↔ χ) ↔ (ψ ↔ (φ ↔ χ)))
5		biass 348	. . 3 ⊢ ((((φ ↔ ψ) ↔ χ) ↔ (ψ ↔ (φ ↔ χ))) ↔ ((φ ↔ ψ) ↔ (χ ↔ (ψ ↔ (φ ↔ χ)))))
6	4, 5	mpbi 199	. 2 ⊢ ((φ ↔ ψ) ↔ (χ ↔ (ψ ↔ (φ ↔ χ))))
7		biass 348	. 2 ⊢ (((χ ↔ ψ) ↔ (φ ↔ χ)) ↔ (χ ↔ (ψ ↔ (φ ↔ χ))))
8	6, 7	bitr4i 243	1 ⊢ ((φ ↔ ψ) ↔ ((χ ↔ ψ) ↔ (φ ↔ χ)))

This proof depends on syntax axioms:   ↔ wb 176

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

This proof depends on definitions:  df-bi 177

This theorem is used by: (None)