Theorem 3jaod 1304

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 1301	. 2 ⊢ (((𝜓 → 𝜒) ∧ (𝜃 → 𝜒) ∧ (𝜏 → 𝜒)) → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
5	1, 2, 3, 4	syl3anc 1238	1 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))

Syntax hints:   → wi 4   ∨ w3o 977

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 709

This theorem depends on definitions:  df-bi 117  df-3or 979  df-3an 980

This theorem is referenced by:  3jaodan  1306  3jaao  1308  issod  4321  nnawordex  6532  exmidontri2or  7244  addlocprlem  7536  nqprloc  7546  ltexprlemrl  7611  aptiprleml  7640  aptiprlemu  7641  elnn0z  9268  zaddcl  9295  zletric  9299  zlelttric  9300  zltnle  9301  zdceq  9330  zdcle  9331  zdclt  9332  nn01to3  9619  xposdif  9884  fzdcel  10042  qletric  10246  qlelttric  10247  qltnle  10248  qdceq  10249  frec2uzlt2d  10406  triap  14862  tridceq  14889