Theorem imdistan 339

Description: Distribution of implication with conjunction.

Assertion

Ref	Expression
imdistan	⊢ ((φ → (ψ → χ)) ↔ ((φ ∧ ψ) → (φ ∧ χ)))

Proof of Theorem imdistan

Step	Hyp	Ref	Expression
1		anc2l 248	. . 3 ⊢ ((φ → (ψ → χ)) → (φ → (ψ → (φ ∧ χ))))
2	1	imp3a 279	. 2 ⊢ ((φ → (ψ → χ)) → ((φ ∧ ψ) → (φ ∧ χ)))
3		pm3.27 260	. . . 4 ⊢ ((φ ∧ χ) → χ)
4	3	syl3 18	. . 3 ⊢ (((φ ∧ ψ) → (φ ∧ χ)) → ((φ ∧ ψ) → χ))
5	4	exp3a 292	. 2 ⊢ (((φ ∧ ψ) → (φ ∧ χ)) → (φ → (ψ → χ)))
6	2, 5	impbi 139	1 ⊢ ((φ → (ψ → χ)) ↔ ((φ ∧ ψ) → (φ ∧ χ)))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196

This theorem is referenced by:  imdistand 342  r19.22 1272  ss2rab 1553

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-an 198

metamath.org