Theorem syland 609

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 416	. . 3 ⊢ (𝜑 → (𝜒 → (𝜃 → 𝜏)))
4	1, 3	syld 47	. 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜏)))
5	4	impd 411	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) → 𝜏))

Syntax hints:   → wi 4   ∧ wa 396

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 208  df-an 397

This theorem is referenced by:  sylani  610  sylan2d  611  syl2and  614  onfununi  8278  fodomfir  9235  lt2add  11633  nn0seqcvgd  16537  1stcelcls  23451  llyidm  23478  filuni  23875  ballotlemimin  34697  rankfilimb  35290  btwnintr  36254  ifscgr  36279  btwnconn1lem12  36333  poimir  38027  cvrntr  39924  goldbachthlem2  48031