Theorem pm4.72 485

Description: Implication in terms of biconditional and disjunction. Theorem *4.72 of [WhiteheadRussell] p. 121.

Assertion

Ref	Expression
pm4.72	⊢ ((φ → ψ) ↔ (ψ ↔ (φ ∨ ψ)))

Proof of Theorem pm4.72

Step	Hyp	Ref	Expression
1		olc 224	. . . 4 ⊢ (ψ → (φ ∨ ψ))
2	1	a1i 7	. . 3 ⊢ ((φ → ψ) → (ψ → (φ ∨ ψ)))
3		pm2.621 211	. . 3 ⊢ ((φ → ψ) → ((φ ∨ ψ) → ψ))
4	2, 3	impbid 397	. 2 ⊢ ((φ → ψ) → (ψ ↔ (φ ∨ ψ)))
5		bi2 131	. . 3 ⊢ ((ψ ↔ (φ ∨ ψ)) → ((φ ∨ ψ) → ψ))
6		pm2.67 231	. . 3 ⊢ (((φ ∨ ψ) → ψ) → (φ → ψ))
7	5, 6	syl 12	. 2 ⊢ ((ψ ↔ (φ ∨ ψ)) → (φ → ψ))
8	4, 7	impbi 139	1 ⊢ ((φ → ψ) ↔ (ψ ↔ (φ ∨ ψ)))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195

This theorem is referenced by:  bigolden 513  ssequn1 1628

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

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

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