Theorem axreplem 5233

Description: Lemma for axrep2 5234 and axrep3 5235. (Contributed by BJ, 6-Aug-2022.)

Assertion

Ref	Expression
axreplem	⊢ (𝑥 = 𝑦 → (∃𝑢(𝜑 → ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑥 ∧ 𝜒))) ↔ ∃𝑢(𝜑 → ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑦 ∧ 𝜒)))))

Distinct variable groups:   𝑥,𝑢   𝑦,𝑢   𝑥,𝑣   𝑦,𝑣   𝑥,𝑤   𝑦,𝑤

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑧, 𝑤, 𝑣, 𝑢)   𝜓(𝑥, 𝑦, 𝑧, 𝑤, 𝑣, 𝑢)   𝜒(𝑥, 𝑦, 𝑧, 𝑤, 𝑣, 𝑢)

Proof of Theorem axreplem

Step	Hyp	Ref	Expression
1		elequ2 2159	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
2	1	anbi1d 643	. . . . . 6 ⊢ (𝑥 = 𝑦 → ((𝑧 ∈ 𝑥 ∧ 𝜒) ↔ (𝑧 ∈ 𝑦 ∧ 𝜒)))
3	2	exbidv 1953	. . . . 5 ⊢ (𝑥 = 𝑦 → (∃𝑤(𝑧 ∈ 𝑥 ∧ 𝜒) ↔ ∃𝑤(𝑧 ∈ 𝑦 ∧ 𝜒)))
4	3	bibi2d 345	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑥 ∧ 𝜒)) ↔ (𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑦 ∧ 𝜒))))
5	4	albidv 1952	. . 3 ⊢ (𝑥 = 𝑦 → (∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑥 ∧ 𝜒)) ↔ ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑦 ∧ 𝜒))))
6	5	imbi2d 343	. 2 ⊢ (𝑥 = 𝑦 → ((𝜑 → ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑥 ∧ 𝜒))) ↔ (𝜑 → ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑦 ∧ 𝜒)))))
7	6	exbidv 1953	1 ⊢ (𝑥 = 𝑦 → (∃𝑢(𝜑 → ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑥 ∧ 𝜒))) ↔ ∃𝑢(𝜑 → ∀𝑣(𝜓 ↔ ∃𝑤(𝑧 ∈ 𝑦 ∧ 𝜒)))))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812

This theorem is used by:  axrep2  5234  axrep3  5235