Theorem syland 604

Description: A syllogism deduction. (Contributed by NM, 15-Dec-2004.)

Hypotheses

Ref	Expression
syland.1	⊢ (𝜑 → (𝜓 → 𝜒))
syland.2	⊢ (𝜑 → ((𝜒 ∧ 𝜃) → 𝜏))

Assertion

Ref	Expression
syland	⊢ (𝜑 → ((𝜓 ∧ 𝜃) → 𝜏))

Proof of Theorem syland

Step	Hyp	Ref	Expression
1		syland.1	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
2		syland.2	. . . 4 ⊢ (𝜑 → ((𝜒 ∧ 𝜃) → 𝜏))
3	2	expd 418	. . 3 ⊢ (𝜑 → (𝜒 → (𝜃 → 𝜏)))
4	1, 3	syld 47	. 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜏)))
5	4	impd 413	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) → 𝜏))

Syntax hints:   → wi 4   ∧ wa 398

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 209  df-an 399

This theorem is referenced by:  sylani  605  sylan2d  606  syl2and  609  onfununi  7972  lt2add  11119  nn0seqcvgd  15908  1stcelcls  22063  llyidm  22090  filuni  22487  ballotlemimin  31758  btwnintr  33475  ifscgr  33500  btwnconn1lem12  33554  poimir  34919  cvrntr  36555  goldbachthlem2  43701