Theorem 3jaod 1283

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 1280	. 2 ⊢ (((𝜓 → 𝜒) ∧ (𝜃 → 𝜒) ∧ (𝜏 → 𝜒)) → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
5	1, 2, 3, 4	syl3anc 1217	1 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))

Syntax hints:   → wi 4   ∨ w3o 962

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 699

This theorem depends on definitions:  df-bi 116  df-3or 964  df-3an 965

This theorem is referenced by:  3jaodan  1285  3jaao  1287  issod  4249  nnawordex  6432  addlocprlem  7367  nqprloc  7377  ltexprlemrl  7442  aptiprleml  7471  aptiprlemu  7472  elnn0z  9091  zaddcl  9118  zletric  9122  zlelttric  9123  zltnle  9124  zdceq  9150  zdcle  9151  zdclt  9152  nn01to3  9436  xposdif  9695  fzdcel  9851  qletric  10052  qlelttric  10053  qltnle  10054  qdceq  10055  frec2uzlt2d  10208  triap  13399