Theorem bj-rexcom4a 32064

Description: Remove from rexcom4a 3199 dependency on ax-ext 2590 and ax-13 2234 (and on df-or 384, df-sb 1868, df-clab 2597, df-cleq 2603, df-clel 2606, df-nfc 2740, df-v 3175). This proof uses only df-rex 2902 on top of first-order logic. (Contributed by BJ, 16-Jun-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-rexcom4a	⊢ (∃𝑥∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜓) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ ∃𝑥𝜓))

Distinct variable groups:   𝑥,𝐴   𝑥,𝑦   𝜑,𝑥

Allowed substitution hints:   𝜑(𝑦)   𝜓(𝑥,𝑦)   𝐴(𝑦)

Proof of Theorem bj-rexcom4a

Step	Hyp	Ref	Expression
1		bj-rexcom4 32063	. 2 ⊢ (∃𝑦 ∈ 𝐴 ∃𝑥(𝜑 ∧ 𝜓) ↔ ∃𝑥∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜓))
2		19.42v 1905	. . 3 ⊢ (∃𝑥(𝜑 ∧ 𝜓) ↔ (𝜑 ∧ ∃𝑥𝜓))
3	2	rexbii 3023	. 2 ⊢ (∃𝑦 ∈ 𝐴 ∃𝑥(𝜑 ∧ 𝜓) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ ∃𝑥𝜓))
4	1, 3	bitr3i 265	1 ⊢ (∃𝑥∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝜓) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ ∃𝑥𝜓))

Syntax hints:   ↔ wb 195   ∧ wa 383  ∃wex 1695  ∃wrex 2897

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-11 2021

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-rex 2902

This theorem is referenced by:  bj-rexcom4bv  32065  bj-rexcom4b  32066