Theorem mpand 656

Description: A deduction based on modus ponens. (Contributed by NM, 12-Dec-2004.) (Proof shortened by Wolf Lammen, 7-Apr-2013.)

Hypotheses

Ref	Expression
mpand.1	⊢ (φ → ψ)
mpand.2	⊢ (φ → ((ψ ∧ χ) → θ))

Assertion

Ref	Expression
mpand	⊢ (φ → (χ → θ))

Proof of Theorem mpand

Step	Hyp	Ref	Expression
1		mpand.1	. 2 ⊢ (φ → ψ)
2		mpand.2	. . 3 ⊢ (φ → ((ψ ∧ χ) → θ))
3	2	ancomsd 440	. 2 ⊢ (φ → ((χ ∧ ψ) → θ))
4	1, 3	mpan2d 655	1 ⊢ (φ → (χ → θ))

This proof depends on syntax axioms:   → wi 4   ∧ wa 358

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  df-an 360

This theorem is used by:  mpani  657  mp2and  660  ecase2d  906  peano5  4410  sfinltfin  4536  vfinspss  4552  fvopab3ig  5388  ovig  5598  ncssfin  6152