Theorem syland 277

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 245	. . 3 ⊢ (𝜑 → (𝜒 → (𝜃 → 𝜏)))
4	1, 3	syld 40	. 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜏)))
5	4	impd 242	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) → 𝜏))

Syntax hints:   → wi 4   ∧ wa 97

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem is referenced by:  sylan2d  278  syl2and  279  sylani  386  nn0seqcvgd  9880