Theorem 3jaoi 1299

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 1171	. 2 ⊢ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓))
5		3jao 1297	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))
6	4, 5	ax-mp 5	1 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)

Syntax hints:   → wi 4   ∨ w3o 973   ∧ w3a 974

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 705

This theorem depends on definitions:  df-bi 116  df-3or 975  df-3an 976

This theorem is referenced by:  3jaoian  1301  3ianorr  1305  acexmidlem1  5853  nndceq  6482  nndcel  6483  znegcl  9247  xrltnr  9740  nltpnft  9775  ngtmnft  9778  xrrebnd  9780  xnegcl  9793  xnegneg  9794  xltnegi  9796  xrpnfdc  9803  xrmnfdc  9804  xnegid  9820  xaddid1  9823  xposdif  9843  prm23lt5  12221  zabsle1  13779