Theorem a2and 859

Description: Deduction distributing a conjunction as embedded antecedent. (Contributed by AV, 25-Oct-2019.) (Proof shortened by Wolf Lammen, 19-Jan-2020.)

Hypotheses

Ref	Expression
a2and.1	⊢ (𝜑 → ((𝜓 ∧ 𝜌) → (𝜏 → 𝜃)))
a2and.2	⊢ (𝜑 → ((𝜓 ∧ 𝜌) → 𝜒))

Assertion

Ref	Expression
a2and	⊢ (𝜑 → (((𝜓 ∧ 𝜒) → 𝜏) → ((𝜓 ∧ 𝜌) → 𝜃)))

Proof of Theorem a2and

Step	Hyp	Ref	Expression
1		a2and.1	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜌) → (𝜏 → 𝜃)))
2		a2and.2	. . . . 5 ⊢ (𝜑 → ((𝜓 ∧ 𝜌) → 𝜒))
3	2	expd 421	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜌 → 𝜒)))
4	3	imdistand 581	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜌) → (𝜓 ∧ 𝜒)))
5		imim1 84	. . . 4 ⊢ (((𝜓 ∧ 𝜒) → 𝜏) → ((𝜏 → 𝜃) → ((𝜓 ∧ 𝜒) → 𝜃)))
6	5	com3l 90	. . 3 ⊢ ((𝜏 → 𝜃) → ((𝜓 ∧ 𝜒) → (((𝜓 ∧ 𝜒) → 𝜏) → 𝜃)))
7	1, 4, 6	syl6c 71	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜌) → (((𝜓 ∧ 𝜒) → 𝜏) → 𝜃)))
8	7	com23 87	1 ⊢ (𝜑 → (((𝜓 ∧ 𝜒) → 𝜏) → ((𝜓 ∧ 𝜌) → 𝜃)))

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:  chnind  18702  telgsumfzs  20090