Theorem redundpim3 39614

Description: Implication of redundancy of proposition. (Contributed by Peter Mazsa, 26-Oct-2022.)

Hypothesis

Ref	Expression
redundpim3.1	⊢ (𝜃 → 𝜒)

Assertion

Ref	Expression
redundpim3	⊢ ( redund (𝜑, 𝜓, 𝜒) → redund (𝜑, 𝜓, 𝜃))

Proof of Theorem redundpim3

Step	Hyp	Ref	Expression
1		anbi1 645	. . . 4 ⊢ (((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒)) → (((𝜑 ∧ 𝜒) ∧ 𝜃) ↔ ((𝜓 ∧ 𝜒) ∧ 𝜃)))
2		redundpim3.1	. . . . . 6 ⊢ (𝜃 → 𝜒)
3	2	pm4.71ri 570	. . . . 5 ⊢ (𝜃 ↔ (𝜒 ∧ 𝜃))
4	3	bianass 655	. . . 4 ⊢ ((𝜑 ∧ 𝜃) ↔ ((𝜑 ∧ 𝜒) ∧ 𝜃))
5	3	bianass 655	. . . 4 ⊢ ((𝜓 ∧ 𝜃) ↔ ((𝜓 ∧ 𝜒) ∧ 𝜃))
6	1, 4, 5	3bitr4g 317	. . 3 ⊢ (((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒)) → ((𝜑 ∧ 𝜃) ↔ (𝜓 ∧ 𝜃)))
7	6	anim2i 629	. 2 ⊢ (((𝜑 → 𝜓) ∧ ((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))) → ((𝜑 → 𝜓) ∧ ((𝜑 ∧ 𝜃) ↔ (𝜓 ∧ 𝜃))))
8		df-redundp 39609	. 2 ⊢ ( redund (𝜑, 𝜓, 𝜒) ↔ ((𝜑 → 𝜓) ∧ ((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))))
9		df-redundp 39609	. 2 ⊢ ( redund (𝜑, 𝜓, 𝜃) ↔ ((𝜑 → 𝜓) ∧ ((𝜑 ∧ 𝜃) ↔ (𝜓 ∧ 𝜃))))
10	7, 8, 9	3imtr4i 295	1 ⊢ ( redund (𝜑, 𝜓, 𝜒) → redund (𝜑, 𝜓, 𝜃))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   redund wredundp 39105

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-redundp 39609

This theorem is used by:  refrelredund2  39620