Theorem mpjaod 370

Description: Eliminate a disjunction in a deduction. (Contributed by Mario Carneiro, 29-May-2016.)

Hypotheses

Ref	Expression
jaod.1	⊢ (φ → (ψ → χ))
jaod.2	⊢ (φ → (θ → χ))
jaod.3	⊢ (φ → (ψ ∨ θ))

Assertion

Ref	Expression
mpjaod	⊢ (φ → χ)

Proof of Theorem mpjaod

Step	Hyp	Ref	Expression
1		jaod.3	. 2 ⊢ (φ → (ψ ∨ θ))
2		jaod.1	. . 3 ⊢ (φ → (ψ → χ))
3		jaod.2	. . 3 ⊢ (φ → (θ → χ))
4	2, 3	jaod 369	. 2 ⊢ (φ → ((ψ ∨ θ) → χ))
5	1, 4	mpd 14	1 ⊢ (φ → χ)

This proof depends on syntax axioms:   → wi 4   ∨ wo 357

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-or 359

This theorem is used by:  ecase2d  906  fnfreclem3  6320