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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 2311 | . . 3 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑)) | |
2 | 1 | anbi1i 431 | . 2 ⊢ ((∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴)) |
3 | df-alsc 10118 | . 2 ⊢ (∀!𝑥 ∈ 𝐴𝜑 ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴)) | |
4 | df-alsi 10117 | . 2 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴)) | |
5 | 2, 3, 4 | 3bitr4ri 202 | 1 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 97 ↔ wb 98 ∀wal 1241 ∃wex 1381 ∈ wcel 1393 ∀wral 2306 ∀!walsi 10115 ∀!walsc 10116 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 |
This theorem depends on definitions: df-bi 110 df-ral 2311 df-alsi 10117 df-alsc 10118 |
This theorem is referenced by: (None) |
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