Theorem 3impexpVD 45837

Description: Virtual deduction proof of 3impexp 1377. The following user's proof is completed by invoking mmj2's unify command and using mmj2's StepSelector to pick all remaining steps of the Metamath proof.

1::	⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   )
2::	⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
3:1,2,?: e10 45676	⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)   )
4:3,?: e1a 45609	⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   ((𝜑 ∧ 𝜓) → (𝜒 → 𝜃))   )
5:4,?: e1a 45609	⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   )
6:5:	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
7::	⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   )
8:7,?: e1a 45609	⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   ((𝜑 ∧ 𝜓) → (𝜒 → 𝜃))   )
9:8,?: e1a 45609	⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)   )
10:2,9,?: e01 45673	⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   )
11:10:	⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃))
qed:6,11,?: e00 45749	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))

(Contributed by Alan Sare, 31-Dec-2011.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
3impexpVD	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))

Proof of Theorem 3impexpVD

Step	Hyp	Ref	Expression
1		idn1 45556	. . . . . 6 ⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   )
2		df-3an 1105	. . . . . 6 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
3		imbi1 350	. . . . . . 7 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒)) → (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)))
4	3	biimpcd 252	. . . . . 6 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → (((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒)) → (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)))
5	1, 2, 4	e10 45676	. . . . 5 ⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)   )
6		pm3.3 454	. . . . 5 ⊢ ((((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃) → ((𝜑 ∧ 𝜓) → (𝜒 → 𝜃)))
7	5, 6	e1a 45609	. . . 4 ⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   ((𝜑 ∧ 𝜓) → (𝜒 → 𝜃))   )
8		pm3.3 454	. . . 4 ⊢ (((𝜑 ∧ 𝜓) → (𝜒 → 𝜃)) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
9	7, 8	e1a 45609	. . 3 ⊢ (   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   ▶   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   )
10	9	in1 45553	. 2 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → (𝜑 → (𝜓 → (𝜒 → 𝜃))))
11		idn1 45556	. . . . . 6 ⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   )
12		pm3.31 455	. . . . . 6 ⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → ((𝜑 ∧ 𝜓) → (𝜒 → 𝜃)))
13	11, 12	e1a 45609	. . . . 5 ⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   ((𝜑 ∧ 𝜓) → (𝜒 → 𝜃))   )
14		pm3.31 455	. . . . 5 ⊢ (((𝜑 ∧ 𝜓) → (𝜒 → 𝜃)) → (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃))
15	13, 14	e1a 45609	. . . 4 ⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)   )
16	3	biimprd 251	. . . 4 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒)) → ((((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)))
17	2, 15, 16	e01 45673	. . 3 ⊢ (   (𝜑 → (𝜓 → (𝜒 → 𝜃)))   ▶   ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)   )
18	17	in1 45553	. 2 ⊢ ((𝜑 → (𝜓 → (𝜒 → 𝜃))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃))
19		impbi 211	. 2 ⊢ ((((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) → (𝜑 → (𝜓 → (𝜒 → 𝜃)))) → (((𝜑 → (𝜓 → (𝜒 → 𝜃))) → ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)) → (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))))
20	10, 18, 19	e00 45749	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ↔ (𝜑 → (𝜓 → (𝜒 → 𝜃))))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∧ w3a 1103

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  df-3an 1105  df-vd1 45552

This theorem is used by: (None)