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  8377  fodomfir  9347  lt2add  11828  nn0seqcvgd  16774  1stcelcls  23810  llyidm  23837  filuni  24234  ballotlemimin  35208  rankfilimb  35812  btwnintr  36962  ifscgr  36987  btwnconn1lem12  37041  poimir  38763  cvrntr  40674  goldbachthlem2  48836