Theorem redundpbi1 35899

Description: Equivalence of redundancy of propositions. (Contributed by Peter Mazsa, 25-Oct-2022.)

Hypothesis

Ref	Expression
redundpbi1.1	⊢ (𝜑 ↔ 𝜃)

Assertion

Ref	Expression
redundpbi1	⊢ ( redund (𝜑, 𝜓, 𝜒) ↔ redund (𝜃, 𝜓, 𝜒))

Proof of Theorem redundpbi1

Step	Hyp	Ref	Expression
1		redundpbi1.1	. . . 4 ⊢ (𝜑 ↔ 𝜃)
2	1	imbi1i 352	. . 3 ⊢ ((𝜑 → 𝜓) ↔ (𝜃 → 𝜓))
3	1	anbi1i 625	. . . 4 ⊢ ((𝜑 ∧ 𝜒) ↔ (𝜃 ∧ 𝜒))
4	3	bibi1i 341	. . 3 ⊢ (((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒)) ↔ ((𝜃 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒)))
5	2, 4	anbi12i 628	. 2 ⊢ (((𝜑 → 𝜓) ∧ ((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))) ↔ ((𝜃 → 𝜓) ∧ ((𝜃 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))))
6		df-redundp 35893	. 2 ⊢ ( redund (𝜑, 𝜓, 𝜒) ↔ ((𝜑 → 𝜓) ∧ ((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))))
7		df-redundp 35893	. 2 ⊢ ( redund (𝜃, 𝜓, 𝜒) ↔ ((𝜃 → 𝜓) ∧ ((𝜃 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))))
8	5, 6, 7	3bitr4i 305	1 ⊢ ( redund (𝜑, 𝜓, 𝜒) ↔ redund (𝜃, 𝜓, 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   redund wredundp 35508

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 209  df-an 399  df-redundp 35893

This theorem is referenced by:  refrelredund3  35905