Theorem 3jaoi 1234

Description: Disjunction of 3 antecedents (inference). (Contributed by NM, 12-Sep-1995.)

Hypotheses

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

Assertion

Ref	Expression
3jaoi	⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)

Proof of Theorem 3jaoi

Step	Hyp	Ref	Expression
1		3jaoi.1	. . 3 ⊢ (𝜑 → 𝜓)
2		3jaoi.2	. . 3 ⊢ (𝜒 → 𝜓)
3		3jaoi.3	. . 3 ⊢ (𝜃 → 𝜓)
4	1, 2, 3	3pm3.2i 1116	. 2 ⊢ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓))
5		3jao 1232	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))
6	4, 5	ax-mp 7	1 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)

Syntax hints:   → wi 4   ∨ w3o 918   ∧ w3a 919

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662

This theorem depends on definitions:  df-bi 115  df-3or 920  df-3an 921

This theorem is referenced by:  3jaoian  1236  3ianorr  1240  acexmidlem1  5528  nndceq  6100  nndcel  6101  znegcl  8382  xrltnr  8855  nltpnft  8884  ngtmnft  8885  xrrebnd  8886  xnegcl  8899  xnegneg  8900  xltnegi  8902