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 415	. . 3 ⊢ (𝜑 → (𝜒 → (𝜃 → 𝜏)))
4	1, 3	syld 47	. 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜏)))
5	4	impd 410	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) → 𝜏))

Syntax hints:   → wi 4   ∧ wa 395

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 207  df-an 396

This theorem is referenced by:  sylani  605  sylan2d  606  syl2and  609  onfununi  8281  fodomfir  9238  lt2add  11635  nn0seqcvgd  16539  1stcelcls  23426  llyidm  23453  filuni  23850  ballotlemimin  34650  rankfilimb  35245  btwnintr  36201  ifscgr  36226  btwnconn1lem12  36280  poimir  37974  cvrntr  39871  goldbachthlem2  48009