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Mirrors > Home > ILE Home > Th. List > Mathboxes > alsconv | GIF version |
Description: There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.) |
Ref | Expression |
---|---|
alsconv | ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-ral 2440 | . . 3 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑)) | |
2 | 1 | anbi1i 454 | . 2 ⊢ ((∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴)) |
3 | df-alsc 13718 | . 2 ⊢ (∀!𝑥 ∈ 𝐴𝜑 ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴)) | |
4 | df-alsi 13717 | . 2 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴)) | |
5 | 2, 3, 4 | 3bitr4ri 212 | 1 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 103 ↔ wb 104 ∀wal 1333 ∃wex 1472 ∈ wcel 2128 ∀wral 2435 ∀!walsi 13715 ∀!walsc 13716 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 |
This theorem depends on definitions: df-bi 116 df-ral 2440 df-alsi 13717 df-alsc 13718 |
This theorem is referenced by: (None) |
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