Theorem 3jaod 1282

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 1279	. 2 ⊢ (((𝜓 → 𝜒) ∧ (𝜃 → 𝜒) ∧ (𝜏 → 𝜒)) → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
5	1, 2, 3, 4	syl3anc 1216	1 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))

Syntax hints:   → wi 4   ∨ w3o 961

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 698

This theorem depends on definitions:  df-bi 116  df-3or 963  df-3an 964

This theorem is referenced by:  3jaodan  1284  3jaao  1286  issod  4241  nnawordex  6424  addlocprlem  7343  nqprloc  7353  ltexprlemrl  7418  aptiprleml  7447  aptiprlemu  7448  elnn0z  9067  zaddcl  9094  zletric  9098  zlelttric  9099  zltnle  9100  zdceq  9126  zdcle  9127  zdclt  9128  nn01to3  9409  xposdif  9665  fzdcel  9820  qletric  10021  qlelttric  10022  qltnle  10023  qdceq  10024  frec2uzlt2d  10177  triap  13224