Theorem ibib 592

Description: Implication in terms of implication and biconditional.

Assertion

Ref	Expression
ibib	⊢ ((φ → ψ) ↔ (φ → (φ ↔ ψ)))

Proof of Theorem ibib

Step	Hyp	Ref	Expression
1		pm3.4 331	. . . . 5 ⊢ ((φ ⋀ ψ) → (φ → ψ))
2		pm3.26 319	. . . . . 6 ⊢ ((φ ⋀ ψ) → φ)
3	2	a1d 12	. . . . 5 ⊢ ((φ ⋀ ψ) → (ψ → φ))
4	1, 3	impbid 518	. . . 4 ⊢ ((φ ⋀ ψ) → (φ ↔ ψ))
5	4	ex 373	. . 3 ⊢ (φ → (ψ → (φ ↔ ψ)))
6		bi1 148	. . . 4 ⊢ ((φ ↔ ψ) → (φ → ψ))
7	6	com12 11	. . 3 ⊢ (φ → ((φ ↔ ψ) → ψ))
8	5, 7	impbid 518	. 2 ⊢ (φ → (ψ ↔ (φ ↔ ψ)))
9	8	pm5.74i 586	1 ⊢ ((φ → ψ) ↔ (φ → (φ ↔ ψ)))

Syntax hints:   → wi 3   ↔ wb 146   ⋀ wa 223

This theorem is referenced by:  ibibr 593  ibd 596  pm5.501 597  zneo 6202

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 147  df-an 225

Copyright terms: Public domain