Theorem 3jaod 1317

Description: Disjunction of 3 antecedents (deduction). (Contributed by NM, 14-Oct-2005.)

Hypotheses

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

Assertion

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

Proof of Theorem 3jaod

Step	Hyp	Ref	Expression
1		3jaod.1	. 2 ⊢ (𝜑 → (𝜓 → 𝜒))
2		3jaod.2	. 2 ⊢ (𝜑 → (𝜃 → 𝜒))
3		3jaod.3	. 2 ⊢ (𝜑 → (𝜏 → 𝜒))
4		3jao 1314	. 2 ⊢ (((𝜓 → 𝜒) ∧ (𝜃 → 𝜒) ∧ (𝜏 → 𝜒)) → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
5	1, 2, 3, 4	syl3anc 1250	1 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))

Syntax hints:   → wi 4   ∨ w3o 980

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711

This theorem depends on definitions:  df-bi 117  df-3or 982  df-3an 983

This theorem is referenced by:  3jaodan  1319  3jaao  1321  issod  4379  nnawordex  6633  exmidontri2or  7384  addlocprlem  7678  nqprloc  7688  ltexprlemrl  7753  aptiprleml  7782  aptiprlemu  7783  elnn0z  9415  zaddcl  9442  zletric  9446  zlelttric  9447  zltnle  9448  zdceq  9478  zdcle  9479  zdclt  9480  nn01to3  9768  xposdif  10034  fzdcel  10192  qletric  10416  qlelttric  10417  qltnle  10418  qdceq  10419  qdclt  10420  frec2uzlt2d  10581  perfectlem2  15557  triap  16140  tridceq  16167