Theorem syland 615

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 421	. . 3 ⊢ (𝜑 → (𝜒 → (𝜃 → 𝜏)))
4	1, 3	syld 48	. 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜏)))
5	4	impd 416	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) → 𝜏))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by:  sylani  616  sylan2d  617  syl2and  620  onfununi  8337  fodomfir  9297  lt2add  11717  nn0seqcvgd  16653  1stcelcls  23655  llyidm  23682  filuni  24079  ballotlemimin  34928  rankfilimb  35521  btwnintr  36532  ifscgr  36557  btwnconn1lem12  36611  poimir  38345  cvrntr  40240  goldbachthlem2  48339