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  7346  nqprloc  7356  ltexprlemrl  7421  aptiprleml  7450  aptiprlemu  7451  elnn0z  9070  zaddcl  9097  zletric  9101  zlelttric  9102  zltnle  9103  zdceq  9129  zdcle  9130  zdclt  9131  nn01to3  9412  xposdif  9668  fzdcel  9823  qletric  10024  qlelttric  10025  qltnle  10026  qdceq  10027  frec2uzlt2d  10180  triap  13227