P. 100 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 9901-10000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	rankpr 9901	The rank of an unordered pair. Part of Exercise 30 of [Enderton] p. 207. (Contributed by NM, 28-Nov-2003.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (rank‘{𝐴, 𝐵}) = suc ((rank‘𝐴) ∪ (rank‘𝐵))
Theorem	rankop 9902	The rank of an ordered pair. Part of Exercise 4 of [Kunen] p. 107. (Contributed by NM, 13-Sep-2006.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (rank‘⟨𝐴, 𝐵⟩) = suc suc ((rank‘𝐴) ∪ (rank‘𝐵))
Theorem	rankung 9903	The rank of the union of two sets. Closed form of rankun 9900. (Contributed by Scott Fenton, 15-Jul-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (rank‘(𝐴 ∪ 𝐵)) = ((rank‘𝐴) ∪ (rank‘𝐵)))
Theorem	ranksng 9904	The rank of a singleton. Closed form of ranksn 9898. (Contributed by Scott Fenton, 15-Jul-2015.)
⊢ (𝐴 ∈ 𝑉 → (rank‘{𝐴}) = suc (rank‘𝐴))
Theorem	r1rankid 9905	Any set is a subset of the hierarchy of its rank. (Contributed by NM, 14-Oct-2003.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ (𝐴 ∈ 𝑉 → 𝐴 ⊆ (𝑅1‘(rank‘𝐴)))
Theorem	rankeq0b 9906	A set is empty iff its rank is empty. (Contributed by Mario Carneiro, 17-Nov-2014.)
⊢ (𝐴 ∈ ∪ (𝑅1 “ On) → (𝐴 = ∅ ↔ (rank‘𝐴) = ∅))
Theorem	rankeq0 9907	A set is empty iff its rank is empty. (Contributed by NM, 18-Sep-2006.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (𝐴 = ∅ ↔ (rank‘𝐴) = ∅)
Theorem	rankr1id 9908	The rank of the hierarchy of an ordinal number is itself. (Contributed by NM, 14-Oct-2003.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ (𝐴 ∈ dom 𝑅1 ↔ (rank‘(𝑅1‘𝐴)) = 𝐴)
Theorem	rankuni 9909	The rank of a union. Part of Exercise 4 of [Kunen] p. 107. (Contributed by NM, 15-Sep-2006.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ (rank‘∪ 𝐴) = ∪ (rank‘𝐴)
Theorem	rankval4b 9910*	The rank of a set is the supremum of the successors of the ranks of its members. Exercise 9.1 of [Jech] p. 72. Also a special case of Theorem 7V(b) of [Enderton] p. 204. This variant of rankval4 9914 does not use Regularity, and so requires the assumption that 𝐴 is in the range of 𝑅1. (Contributed by BTernaryTau, 19-Jan-2026.)
⊢ (𝐴 ∈ ∪ (𝑅1 “ On) → (rank‘𝐴) = ∪ 𝑥 ∈ 𝐴 suc (rank‘𝑥))
Theorem	rankr1b 9911	A relationship between rank and 𝑅1. See rankr1a 9878 for the membership version. (Contributed by NM, 15-Sep-2006.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (𝐵 ∈ On → (𝐴 ⊆ (𝑅1‘𝐵) ↔ (rank‘𝐴) ⊆ 𝐵))
Theorem	ranksuc 9912	The rank of a successor. (Contributed by NM, 18-Sep-2006.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (rank‘suc 𝐴) = suc (rank‘𝐴)
Theorem	rankuniss 9913	Upper bound of the rank of a union. Part of Exercise 30 of [Enderton] p. 207. (Contributed by NM, 30-Nov-2003.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (rank‘∪ 𝐴) ⊆ (rank‘𝐴)
Theorem	rankval4 9914*	The rank of a set is the supremum of the successors of the ranks of its members. Exercise 9.1 of [Jech] p. 72. Also a special case of Theorem 7V(b) of [Enderton] p. 204. (Contributed by NM, 12-Oct-2003.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (rank‘𝐴) = ∪ 𝑥 ∈ 𝐴 suc (rank‘𝑥)
Theorem	rankbnd 9915*	The rank of a set is bounded by a bound for the successor of its members. (Contributed by NM, 18-Sep-2006.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (∀𝑥 ∈ 𝐴 suc (rank‘𝑥) ⊆ 𝐵 ↔ (rank‘𝐴) ⊆ 𝐵)
Theorem	rankbnd2 9916*	The rank of a set is bounded by the successor of a bound for its members. (Contributed by NM, 15-Sep-2006.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (𝐵 ∈ On → (∀𝑥 ∈ 𝐴 (rank‘𝑥) ⊆ 𝐵 ↔ (rank‘𝐴) ⊆ suc 𝐵))
Theorem	rankc1 9917*	A relationship that can be used for computation of rank. (Contributed by NM, 16-Sep-2006.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (∀𝑥 ∈ 𝐴 (rank‘𝑥) ∈ (rank‘∪ 𝐴) ↔ (rank‘𝐴) = (rank‘∪ 𝐴))
Theorem	rankc2 9918*	A relationship that can be used for computation of rank. (Contributed by NM, 16-Sep-2006.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (∃𝑥 ∈ 𝐴 (rank‘𝑥) = (rank‘∪ 𝐴) → (rank‘𝐴) = suc (rank‘∪ 𝐴))
Theorem	rankelun 9919	Rank membership is inherited by union. (Contributed by NM, 18-Sep-2006.) (Proof shortened by Mario Carneiro, 17-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    &   ⊢ 𝐶 ∈ V    &   ⊢ 𝐷 ∈ V    ⇒   ⊢ (((rank‘𝐴) ∈ (rank‘𝐶) ∧ (rank‘𝐵) ∈ (rank‘𝐷)) → (rank‘(𝐴 ∪ 𝐵)) ∈ (rank‘(𝐶 ∪ 𝐷)))
Theorem	rankelpr 9920	Rank membership is inherited by unordered pairs. (Contributed by NM, 18-Sep-2006.) (Revised by Mario Carneiro, 17-Nov-2014.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    &   ⊢ 𝐶 ∈ V    &   ⊢ 𝐷 ∈ V    ⇒   ⊢ (((rank‘𝐴) ∈ (rank‘𝐶) ∧ (rank‘𝐵) ∈ (rank‘𝐷)) → (rank‘{𝐴, 𝐵}) ∈ (rank‘{𝐶, 𝐷}))
Theorem	rankelop 9921	Rank membership is inherited by ordered pairs. (Contributed by NM, 18-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    &   ⊢ 𝐶 ∈ V    &   ⊢ 𝐷 ∈ V    ⇒   ⊢ (((rank‘𝐴) ∈ (rank‘𝐶) ∧ (rank‘𝐵) ∈ (rank‘𝐷)) → (rank‘⟨𝐴, 𝐵⟩) ∈ (rank‘⟨𝐶, 𝐷⟩))
Theorem	rankxpl 9922	A lower bound on the rank of a Cartesian product. (Contributed by NM, 18-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((𝐴 × 𝐵) ≠ ∅ → (rank‘(𝐴 ∪ 𝐵)) ⊆ (rank‘(𝐴 × 𝐵)))
Theorem	rankxpu 9923	An upper bound on the rank of a Cartesian product. (Contributed by NM, 18-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (rank‘(𝐴 × 𝐵)) ⊆ suc suc (rank‘(𝐴 ∪ 𝐵))
Theorem	rankfu 9924	An upper bound on the rank of a function. (Contributed by Gérard Lang, 5-Aug-2018.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (𝐹:𝐴⟶𝐵 → (rank‘𝐹) ⊆ suc suc (rank‘(𝐴 ∪ 𝐵)))
Theorem	rankmapu 9925	An upper bound on the rank of set exponentiation. (Contributed by Gérard Lang, 5-Aug-2018.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (rank‘(𝐴 ↑m 𝐵)) ⊆ suc suc suc (rank‘(𝐴 ∪ 𝐵))
Theorem	rankxplim 9926	The rank of a Cartesian product when the rank of the union of its arguments is a limit ordinal. Part of Exercise 4 of [Kunen] p. 107. See rankxpsuc 9929 for the successor case. (Contributed by NM, 19-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ ((Lim (rank‘(𝐴 ∪ 𝐵)) ∧ (𝐴 × 𝐵) ≠ ∅) → (rank‘(𝐴 × 𝐵)) = (rank‘(𝐴 ∪ 𝐵)))
Theorem	rankxplim2 9927	If the rank of a Cartesian product is a limit ordinal, so is the rank of the union of its arguments. (Contributed by NM, 19-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (Lim (rank‘(𝐴 × 𝐵)) → Lim (rank‘(𝐴 ∪ 𝐵)))
Theorem	rankxplim3 9928	The rank of a Cartesian product is a limit ordinal iff its union is. (Contributed by NM, 19-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (Lim (rank‘(𝐴 × 𝐵)) ↔ Lim ∪ (rank‘(𝐴 × 𝐵)))
Theorem	rankxpsuc 9929	The rank of a Cartesian product when the rank of the union of its arguments is a successor ordinal. Part of Exercise 4 of [Kunen] p. 107. See rankxplim 9926 for the limit ordinal case. (Contributed by NM, 19-Sep-2006.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 ∈ V    ⇒   ⊢ (((rank‘(𝐴 ∪ 𝐵)) = suc 𝐶 ∧ (𝐴 × 𝐵) ≠ ∅) → (rank‘(𝐴 × 𝐵)) = suc suc (rank‘(𝐴 ∪ 𝐵)))
Theorem	tcwf 9930	The transitive closure function is well-founded if its argument is. (Contributed by Mario Carneiro, 23-Jun-2013.)
⊢ (𝐴 ∈ ∪ (𝑅1 “ On) → (TC‘𝐴) ∈ ∪ (𝑅1 “ On))
Theorem	tcrank 9931	This theorem expresses two different facts from the two subset implications in this equality. In the forward direction, it says that the transitive closure has members of every rank below 𝐴. Stated another way, to construct a set at a given rank, you have to climb the entire hierarchy of ordinals below (rank‘𝐴), constructing at least one set at each level in order to move up the ranks. In the reverse direction, it says that every member of (TC‘𝐴) has a rank below the rank of 𝐴, since intuitively it contains only the members of 𝐴 and the members of those and so on, but nothing "bigger" than 𝐴. (Contributed by Mario Carneiro, 23-Jun-2013.)
⊢ (𝐴 ∈ ∪ (𝑅1 “ On) → (rank‘𝐴) = (rank “ (TC‘𝐴)))
Theorem	rankfilimbi 9932*	If all elements of a finite well-founded set have a rank less than a limit ordinal, then the rank of that set is also less than the limit ordinal. (Contributed by BTernaryTau, 19-Jan-2026.)
⊢ (((𝐴 ∈ Fin ∧ 𝐴 ∈ ∪ (𝑅1 “ On)) ∧ (∀𝑥 ∈ 𝐴 (rank‘𝑥) ∈ 𝐵 ∧ Lim 𝐵)) → (rank‘𝐴) ∈ 𝐵)
Theorem	r1filimi 9933*	If all elements of a finite set appear in the cumulative hierarchy prior to a limit ordinal, then that set also appears in the cumulative hierarchy prior to the limit ordinal. (Contributed by BTernaryTau, 19-Jan-2026.)
⊢ ((𝐴 ∈ Fin ∧ ∀𝑥 ∈ 𝐴 𝑥 ∈ ∪ (𝑅1 “ 𝐵) ∧ Lim 𝐵) → 𝐴 ∈ ∪ (𝑅1 “ 𝐵))
2.6.10  Hereditarily finite sets
Syntax	chf 9934	Extend class notation with the class of hereditarily finite sets.
class HF
Definition	df-hf 9935	Define the class of sets belonging to the finite stages of the cumulative hierarchy of sets. This is the class of sets of finite rank by elhf2 9940. They are called the hereditarily finite sets since they are the finite sets whose members are hereditarily finite, as proved in elhf3 9943. (Contributed by Scott Fenton, 9-Jul-2015.)
⊢ HF = ∪ (𝑅1 “ ω)
Theorem	dfhf2 9936	Alternate definition of the class of hereditarily finite sets as the value of the cumulative hierarchy of sets function at ω. This characterization is simpler but requires the axiom of infinity to hold. (Contributed by BTernaryTau, 25-Jan-2026.) Restate using the defined HF symbol. (Revised by Eric Schmidt, 24-Sep-2026.)
⊢ HF = (𝑅1‘ω)
Theorem	elhf 9937*	Membership in the hereditarily finite sets. (Contributed by Scott Fenton, 9-Jul-2015.) Reduce axiom usage and shorten proof. (Revised by BJ, 27-Sep-2026.)
⊢ (𝐴 ∈ HF ↔ ∃𝑥 ∈ ω 𝐴 ∈ (𝑅1‘𝑥))
Theorem	elhfOLD 9938*	Obsolete version of elhf 9937 as of 27-Sep-2026. (Contributed by Scott Fenton, 9-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ∈ HF ↔ ∃𝑥 ∈ ω 𝐴 ∈ (𝑅1‘𝑥))
Theorem	hffi 9939	Hereditarily finite sets are finite sets. (Contributed by BTernaryTau, 30-Dec-2025.) Restate using the defined HF symbol. (Revised by Eric Schmidt, 8-Sep-2026.)
⊢ (𝐴 ∈ HF → 𝐴 ∈ Fin)
Theorem	elhf2 9940	Alternate form of membership in the hereditarily finite sets. (Contributed by Scott Fenton, 13-Jul-2015.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (𝐴 ∈ HF ↔ (rank‘𝐴) ∈ ω)
Theorem	elhf2g 9941	Hereditarily finiteness via rank. Closed form of elhf2 9940. (Contributed by Scott Fenton, 15-Jul-2015.)
⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ HF ↔ (rank‘𝐴) ∈ ω))
Theorem	elhf4 9942*	A set is hereditarily finite iff it is finite and all of its elements are hereditarily finite. (Contributed by BTernaryTau, 19-Jan-2026.) Use HF. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ (𝐴 ∈ HF ↔ (𝐴 ∈ Fin ∧ ∀𝑥 ∈ 𝐴 𝑥 ∈ HF ))
Theorem	elhf3 9943	A set is hereditarily finite if and only if it is finite and all its members are hereditarily finite. (Contributed by Eric Schmidt, 8-Sep-2026.) Avoid ax-reg 9616, ax-inf2 9672. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ (𝐴 ∈ HF ↔ (𝐴 ∈ Fin ∧ 𝐴 ⊆ HF ))
Theorem	hfelhf 9944	Any member of a hereditarily finite set is itself a hereditarily finite set. (Contributed by Scott Fenton, 16-Jul-2015.) Avoid ax-reg 9616, ax-inf2 9672. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ HF ) → 𝐴 ∈ HF )
Theorem	hfsshf 9945	Any subset of a hereditarily finite set is itself a hereditarily finite set. (Contributed by BTernaryTau, 17-Sep-2026.)
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ HF ) → 𝐴 ∈ HF )
Theorem	hfelhfOLD 9946	Obsolete version of elhf3 9943 as of 17-Sep-2026. (Contributed by Scott Fenton, 16-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ HF ) → 𝐴 ∈ HF )
Theorem	0hf 9947	The empty set is a hereditarily finite set. (Contributed by Scott Fenton, 9-Jul-2015.)
⊢ ∅ ∈ HF
Theorem	hfun 9948	The union of two hereditarily finite sets is a hereditarily finite set. (Contributed by Scott Fenton, 15-Jul-2015.) Avoid ax-reg 9616, ax-inf2 9672. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ ((𝐴 ∈ HF ∧ 𝐵 ∈ HF ) → (𝐴 ∪ 𝐵) ∈ HF )
Theorem	hfunOLD 9949	Obsolete version of hfun 9948 as of 17-Sep-2026. (Contributed by Scott Fenton, 15-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((𝐴 ∈ HF ∧ 𝐵 ∈ HF ) → (𝐴 ∪ 𝐵) ∈ HF )
Theorem	hfsn 9950	The singleton of a hereditarily finite set is a hereditarily finite set. (Contributed by Scott Fenton, 15-Jul-2015.) Avoid ax-reg 9616, ax-inf2 9672. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ (𝐴 ∈ HF → {𝐴} ∈ HF )
Theorem	hfsnOLD 9951	Obsolete version of hfsn 9950 as of 17-Sep-2026. (Contributed by Scott Fenton, 15-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ∈ HF → {𝐴} ∈ HF )
Theorem	hfadj 9952	Adjoining one hereditarily finite set to a hereditarily finite set results in a hereditarily finite set. (Contributed by Scott Fenton, 15-Jul-2015.)
⊢ ((𝐴 ∈ HF ∧ 𝐵 ∈ HF ) → (𝐴 ∪ {𝐵}) ∈ HF )
Theorem	elhf3OLD 9953	Obsolete version of elhf3 9943 as of 17-Sep-2026. (Contributed by Eric Schmidt, 8-Sep-2026.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ∈ HF ↔ (𝐴 ∈ Fin ∧ 𝐴 ⊆ HF ))
Theorem	hfuni 9954	The union of a hereditarily finite set is hereditarily finite. (Contributed by Scott Fenton, 16-Jul-2015.) Avoid ax-reg 9616, ax-inf2 9672. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ (𝐴 ∈ HF → ∪ 𝐴 ∈ HF )
Theorem	hfuniOLD 9955	Obsolete version of hfuni 9954 as of 17-Sep-2026. (Contributed by Scott Fenton, 16-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ∈ HF → ∪ 𝐴 ∈ HF )
Theorem	hfpw 9956	The power class of a hereditarily finite set is hereditarily finite. (Contributed by Scott Fenton, 16-Jul-2015.) Avoid ax-reg 9616, ax-inf2 9672. (Revised by BTernaryTau, 17-Sep-2026.)
⊢ (𝐴 ∈ HF → 𝒫 𝐴 ∈ HF )
Theorem	hfpwOLD 9957	Obsolete version of hfpw 9956 as of 17-Sep-2026. (Contributed by Scott Fenton, 16-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ∈ HF → 𝒫 𝐴 ∈ HF )
2.6.11  Scott's trick; collection principle; Hilbert's epsilon
Syntax	cscott 9958	Extend class notation with the Scott's trick operation.
class Scott 𝐴
Definition	df-scott 9959*	Define an operation for Scott's trick, expressed as in Equation 9.3 of [Jech] p. 72. Scott's trick collects all sets that have a certain property and are of the smallest possible rank. The resulting collection is guaranteed to be a set (see scottex 9963). Under this definition, the property in question is represented by membership in a class. See scottab 9970 for the case where a wff (meta)variable is used to represent the property. (Contributed by Rohan Ridenour, 9-Aug-2023.)
⊢ Scott 𝐴 = {𝑥 ∈ 𝐴 ∣ ∀𝑦 ∈ 𝐴 (rank‘𝑥) ⊆ (rank‘𝑦)}
Theorem	scotteqd 9960	Equality theorem for the Scott operation. (Contributed by Rohan Ridenour, 9-Aug-2023.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → Scott 𝐴 = Scott 𝐵)
Theorem	scotteq 9961	Closed form of scotteqd 9960. (Contributed by Rohan Ridenour, 9-Aug-2023.)
⊢ (𝐴 = 𝐵 → Scott 𝐴 = Scott 𝐵)
Theorem	nfscott 9962	Bound-variable hypothesis builder for the Scott operation. (Contributed by Rohan Ridenour, 11-Aug-2023.)
⊢ Ⅎ𝑥𝐴    ⇒   ⊢ Ⅎ𝑥Scott 𝐴
Theorem	scottex 9963	Scott's trick produces a set. (Contributed by NM, 13-Oct-2003.) Use the Scott operation. (Revised by BTernaryTau, 18-Jul-2026.)
⊢ Scott 𝐴 ∈ V
Theorem	scottexOLD 9964*	Obsolete version of scottex 9963 as of 18-Jul-2026. (Contributed by NM, 13-Oct-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ {𝑥 ∈ 𝐴 ∣ ∀𝑦 ∈ 𝐴 (rank‘𝑥) ⊆ (rank‘𝑦)} ∈ V
Theorem	scottss 9965	Scott's trick produces a subset of the input class. (Contributed by Rohan Ridenour, 11-Aug-2023.)
⊢ Scott 𝐴 ⊆ 𝐴
Theorem	scott0 9966	Applying Scott's trick to the empty set leaves it unchanged. (Contributed by BTernaryTau, 3-Jul-2026.)
⊢ Scott ∅ = ∅
Theorem	scott0b 9967	Applying Scott's trick yields the empty set iff it was applied to the empty set. (Contributed by NM, 15-Oct-2003.) Use the Scott operation. (Revised by BTernaryTau, 19-Jul-2026.)
⊢ (𝐴 = ∅ ↔ Scott 𝐴 = ∅)
Theorem	scott0OLD 9968*	Obsolete version of scott0b 9967 as of 19-Jul-2026. (Contributed by NM, 15-Oct-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 = ∅ ↔ {𝑥 ∈ 𝐴 ∣ ∀𝑦 ∈ 𝐴 (rank‘𝑥) ⊆ (rank‘𝑦)} = ∅)
Theorem	scottabf 9969*	Value of the Scott operation at a class abstraction. Variant of scottab 9970 with a nonfreeness hypothesis instead of a disjoint variable condition. (Contributed by Rohan Ridenour, 14-Aug-2023.)
⊢ Ⅎ𝑥𝜓    &   ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))    ⇒   ⊢ Scott {𝑥 ∣ 𝜑} = {𝑥 ∣ (𝜑 ∧ ∀𝑦(𝜓 → (rank‘𝑥) ⊆ (rank‘𝑦)))}
Theorem	scottab 9970*	Value of the Scott operation at a class abstraction. (Contributed by Rohan Ridenour, 14-Aug-2023.)
⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))    ⇒   ⊢ Scott {𝑥 ∣ 𝜑} = {𝑥 ∣ (𝜑 ∧ ∀𝑦(𝜓 → (rank‘𝑥) ⊆ (rank‘𝑦)))}
Theorem	scottabes 9971*	Value of the Scott operation at a class abstraction. Variant of scottab 9970 using explicit substitution. (Contributed by Rohan Ridenour, 14-Aug-2023.)
⊢ Scott {𝑥 ∣ 𝜑} = {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))}
Theorem	elscottab 9972*	An element of the output of the Scott operation applied to a class abstraction satisfies the class abstraction's predicate. (Contributed by Rohan Ridenour, 14-Aug-2023.)
⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))    ⇒   ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝜓)
Theorem	scottexsOLD 9973*	Obsolete theorem as of 27-Jul-2026. Use scottex 9963 with a class abstraction instead. (Contributed by NM, 13-Oct-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ∈ V
Theorem	scott0bs 9974	Theorem scheme version of scott0b 9967. The collection of all 𝑥 of minimum rank such that 𝜑(𝑥) is true, is not empty iff there is an 𝑥 such that 𝜑(𝑥) holds. (Contributed by NM, 13-Oct-2003.) Use the Scott operation. (Revised by BTernaryTau, 22-Jul-2026.)
⊢ (∃𝑥𝜑 ↔ Scott {𝑥 ∣ 𝜑} ≠ ∅)
Theorem	scott0bsOLD 9975*	Obsolete version of scott0bs 9974 as of 19-Jul-2026. (Contributed by NM, 13-Oct-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (∃𝑥𝜑 ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ≠ ∅)
Theorem	scottex2OLD 9976	Obsolete version of scottex 9963 as of 18-Jul-2026. (Contributed by Rohan Ridenour, 9-Aug-2023.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ Scott 𝐴 ∈ V
Theorem	scotteld 9977*	The Scott operation sends inhabited classes to inhabited sets. (Contributed by Rohan Ridenour, 11-Aug-2023.)
⊢ (𝜑 → ∃𝑥 𝑥 ∈ 𝐴)    ⇒   ⊢ (𝜑 → ∃𝑥 𝑥 ∈ Scott 𝐴)
Theorem	scottelrankd 9978	Property of a Scott's trick set. (Contributed by Rohan Ridenour, 11-Aug-2023.)
⊢ (𝜑 → 𝐵 ∈ Scott 𝐴)    &   ⊢ (𝜑 → 𝐶 ∈ Scott 𝐴)    ⇒   ⊢ (𝜑 → (rank‘𝐵) ⊆ (rank‘𝐶))
Theorem	scottrankd 9979	Rank of a nonempty Scott's trick set. (Contributed by Rohan Ridenour, 11-Aug-2023.)
⊢ (𝜑 → 𝐵 ∈ Scott 𝐴)    ⇒   ⊢ (𝜑 → (rank‘Scott 𝐴) = suc (rank‘𝐵))
Theorem	cplem1 9980	Lemma for the Collection Principle cp 9984. (Contributed by NM, 17-Oct-2003.) Use the Scott operation. (Revised by BTernaryTau, 19-Jul-2026.)
⊢ 𝐶 = ∪ 𝑥 ∈ 𝐴 Scott 𝐵    ⇒   ⊢ ∀𝑥 ∈ 𝐴 (𝐵 ≠ ∅ → (𝐵 ∩ 𝐶) ≠ ∅)
Theorem	cplem1OLD 9981*	Obsolete version of cplem1 9980 as of 19-Jul-2026. (Contributed by NM, 17-Oct-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝐶 = {𝑦 ∈ 𝐵 ∣ ∀𝑧 ∈ 𝐵 (rank‘𝑦) ⊆ (rank‘𝑧)}    &   ⊢ 𝐷 = ∪ 𝑥 ∈ 𝐴 𝐶    ⇒   ⊢ ∀𝑥 ∈ 𝐴 (𝐵 ≠ ∅ → (𝐵 ∩ 𝐷) ≠ ∅)
Theorem	cplem2 9982*	Lemma for the Collection Principle cp 9984. (Contributed by NM, 17-Oct-2003.) Use the Scott operation. (Revised by BTernaryTau, 19-Jul-2026.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ∃𝑦∀𝑥 ∈ 𝐴 (𝐵 ≠ ∅ → (𝐵 ∩ 𝑦) ≠ ∅)
Theorem	cplem2OLD 9983*	Obsolete version of cplem2 9982 as of 19-Jul-2026. (Contributed by NM, 17-Oct-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ∃𝑦∀𝑥 ∈ 𝐴 (𝐵 ≠ ∅ → (𝐵 ∩ 𝑦) ≠ ∅)
Theorem	cp 9984*	Collection Principle. This remarkable theorem scheme is in effect a very strong generalization of the Axiom of Replacement. The proof makes use of Scott's trick df-scott 9959 to collapse a proper class into a set of minimum rank. The wff 𝜑 can be thought of as 𝜑(𝑥, 𝑦). Scheme "Collection Principle" of [Jech] p. 72. (Contributed by NM, 17-Oct-2003.)
⊢ ∃𝑤∀𝑥 ∈ 𝑧 (∃𝑦𝜑 → ∃𝑦 ∈ 𝑤 𝜑)
Theorem	bnd 9985*	A very strong generalization of the Axiom of Replacement (compare zfrep6 5246), derived from the Collection Principle cp 9984. Its strength lies in the rather profound fact that 𝜑(𝑥, 𝑦) does not have to be a "function-like" wff, as it does in the standard Axiom of Replacement. This theorem is sometimes called the Boundedness Axiom. (Contributed by NM, 17-Oct-2004.)
⊢ (∀𝑥 ∈ 𝑧 ∃𝑦𝜑 → ∃𝑤∀𝑥 ∈ 𝑧 ∃𝑦 ∈ 𝑤 𝜑)
Theorem	bnd2 9986*	A variant of the Boundedness Axiom bnd 9985 that picks a subset 𝑧 out of a possibly proper class 𝐵 in which a property is true. (Contributed by NM, 4-Feb-2004.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 → ∃𝑧(𝑧 ⊆ 𝐵 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑧 𝜑))
Theorem	kardex 9987	The collection of all sets equinumerous to a set 𝐴 and having the least possible rank is a set. This is the part of the justification of the definition of kard of [Enderton] p. 222. (Contributed by NM, 14-Dec-2003.) Use the Scott operation. (Revised by BTernaryTau, 19-Jul-2026.)
⊢ Scott {𝑥 ∣ 𝑥 ≈ 𝐴} ∈ V
Theorem	kardexOLD 9988*	Obsolete version of kardex 9987 as of 19-Jul-2026. (Contributed by NM, 14-Dec-2003.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ {𝑥 ∣ (𝑥 ≈ 𝐴 ∧ ∀𝑦(𝑦 ≈ 𝐴 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ∈ V
Theorem	karden 9989*	If we allow the Axiom of Regularity, we can avoid the Axiom of Choice by defining the cardinal number of a set as the set of all sets equinumerous to it and having the least possible rank. This theorem proves the equinumerosity relationship for this definition (compare carden 10665). The hypotheses correspond to the definition of kard of [Enderton] p. 222 (which we don't define separately since currently we do not use it elsewhere). This theorem along with kardex 9987 justify the definition of kard. The restriction to the least rank prevents the proper class that would result from {𝑥 ∣ 𝑥 ≈ 𝐴}. (Contributed by NM, 18-Dec-2003.) (Revised by AV, 12-Jul-2022.) Use the Scott operation. (Revised by BTernaryTau, 19-Jul-2026.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐶 = Scott {𝑥 ∣ 𝑥 ≈ 𝐴}    &   ⊢ 𝐷 = Scott {𝑥 ∣ 𝑥 ≈ 𝐵}    ⇒   ⊢ (𝐶 = 𝐷 ↔ 𝐴 ≈ 𝐵)
Theorem	kardenOLD 9990*	Obsolete version of karden 9989 as of 19-Jul-2026. (Contributed by NM, 18-Dec-2003.) (Revised by AV, 12-Jul-2022.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐶 = {𝑥 ∣ (𝑥 ≈ 𝐴 ∧ ∀𝑦(𝑦 ≈ 𝐴 → (rank‘𝑥) ⊆ (rank‘𝑦)))}    &   ⊢ 𝐷 = {𝑥 ∣ (𝑥 ≈ 𝐵 ∧ ∀𝑦(𝑦 ≈ 𝐵 → (rank‘𝑥) ⊆ (rank‘𝑦)))}    ⇒   ⊢ (𝐶 = 𝐷 ↔ 𝐴 ≈ 𝐵)
Theorem	htalem 9991*	Lemma for defining an emulation of Hilbert's epsilon. Hilbert's epsilon is described at http://plato.stanford.edu/entries/epsilon-calculus/. This theorem is equivalent to Hilbert's "transfinite axiom", described on that page, with the additional 𝑅 We 𝐴 antecedent. The element 𝐵 is the epsilon that the theorem emulates. (Contributed by NM, 11-Mar-2004.) (Revised by Mario Carneiro, 25-Jun-2015.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐵 = (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑦𝑅𝑥)    ⇒   ⊢ ((𝑅 We 𝐴 ∧ 𝐴 ≠ ∅) → 𝐵 ∈ 𝐴)
Theorem	hta 9992*	A ZFC emulation of Hilbert's transfinite axiom. The set 𝐵 has the properties of Hilbert's epsilon, except that it also depends on a well-ordering 𝑅. This theorem arose from discussions with Raph Levien on 5-Mar-2004 about translating the HOL proof language, which uses Hilbert's epsilon. See https://us.metamath.org/downloads/choice.txt (copy of obsolete link http://ghilbert.org/choice.txt) and https://us.metamath.org/downloads/megillaward2005he.pdf. Hilbert's epsilon is described at http://plato.stanford.edu/entries/epsilon-calculus/. This theorem differs from Hilbert's transfinite axiom described on that page in that it requires 𝑅 We 𝐴 as an antecedent. Class 𝐴 collects the sets of the least rank for which 𝜑(𝑥) is true. Class 𝐵, which emulates Hilbert's epsilon, is the minimum element in a well-ordering 𝑅 on 𝐴. If a well-ordering 𝑅 on 𝐴 can be expressed in a closed form, as might be the case if we are working with say natural numbers, we can eliminate the antecedent with modus ponens, giving us the exact equivalent of Hilbert's transfinite axiom. Otherwise, we replace 𝑅 with a dummy setvar variable, say 𝑤, and attach 𝑤 We 𝐴 as an antecedent in each step of the ZFC version of the HOL proof until the epsilon is eliminated. At that point, 𝐵 (which will have 𝑤 as a free variable) will no longer be present, and we can eliminate 𝑤 We 𝐴 by applying exlimiv 1965 and weth 10603, using scottex 9963 to establish the existence of 𝐴. For a version of this theorem scheme using class (meta)variables instead of wff (meta)variables, see htalem 9991. (Contributed by NM, 11-Mar-2004.) (Revised by Mario Carneiro, 25-Jun-2015.) Use the Scott operation. (Revised by BTernaryTau, 22-Jul-2026.)
⊢ 𝐴 = Scott {𝑥 ∣ 𝜑}    &   ⊢ 𝐵 = (℩𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ¬ 𝑧𝑅𝑦)    ⇒   ⊢ (𝑅 We 𝐴 → (𝜑 → [𝐵 / 𝑥]𝜑))
Theorem	htaOLD 9993*	Obsolete version of hta 9992 as of 22-Jul-2026. (Contributed by NM, 11-Mar-2004.) (Revised by Mario Carneiro, 25-Jun-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝐴 = {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))}    &   ⊢ 𝐵 = (℩𝑧 ∈ 𝐴 ∀𝑤 ∈ 𝐴 ¬ 𝑤𝑅𝑧)    ⇒   ⊢ (𝑅 We 𝐴 → (𝜑 → [𝐵 / 𝑥]𝜑))
2.6.12  Set Recursion
2.6.12.1  Basic Properties of Set Recursion Symbols in this section: All the symbols used in the definition of setrecs(𝐹) are explained in the comment of df-setrecs 9995. The class 𝑌 is explained in the comment of setrec1lem1 9996. Glossaries of symbols used in individual proofs, or used differently in different proofs, are in the comments of those proofs.
Syntax	csetrecs 9994	Extend class notation to include a set defined by transfinite recursion.
class setrecs(𝐹)
Definition	df-setrecs 9995*	Define a class setrecs(𝐹) by transfinite recursion, where (𝐹‘𝑥) is the set of new elements to add to the class given the set 𝑥 of elements in the class so far. We do not need a base case, because we can start with the empty set, which is vacuously a subset of setrecs(𝐹). The goal of this definition is to construct a class fulfilling Theorems setrec1 10002 and setrec2v 10008, which give a more intuitive idea of the meaning of setrecs. Unlike wrecs, setrecs is well-defined for any 𝐹 and meaningful for any function 𝐹. For example, see Theorem onsetrec 51091 for how the class On is defined recursively using the successor function. The definition works by building subsets of the desired class and taking the union of those subsets. To find such a collection of subsets, consider an arbitrary set 𝑧, and consider the result when applying 𝐹 to any subset 𝑤 ⊆ 𝑧. Remember that 𝐹 can be any function, and in general we are interested in functions that give outputs that are larger than their inputs, so we have no reason to expect the outputs to be within 𝑧. However, if we restrict the domain of 𝐹 to a given set 𝑦, the resulting range will be a set. Therefore, with this restricted 𝐹, it makes sense to consider sets 𝑧 that are closed under 𝐹 applied to its subsets. Now we can test whether a given set 𝑦 is recursively generated by 𝐹. If every set 𝑧 that is closed under 𝐹 contains 𝑦, that means that every member of 𝑦 must eventually be generated by 𝐹. On the other hand, if some such 𝑧 does not contain a certain element of 𝑦, then that element can be avoided even if we apply 𝐹 in every possible way to previously generated elements. Note that such an omitted element might be eventually recursively generated by 𝐹, but not through the elements of 𝑦. In this case, 𝑦 would fail the condition in the definition, but the omitted element would still be included in some larger 𝑦. For example, if 𝐹 is the successor function, the set {∅, 2o} would fail the condition since 2o is not an element of the successor of ∅ or {∅}. Remember that we are applying 𝐹 to subsets of 𝑦, not elements of 𝑦. In fact, even the set {1o} fails the condition, since the only subset of previously generated elements is ∅, and suc ∅ does not have 1o as an element. However, we can let 𝑦 be any ordinal, since each of its elements is generated by starting with ∅ and repeatedly applying the successor function. A similar definition I initially used for setrecs(𝐹) was setrecs(𝐹) = ∪ ran recs((𝑔 ∈ V ↦ (𝐹‘∪ ran 𝑔))). I had initially tried and failed to find an elementary definition, and I had proven theorems analogous to setrec1 10002 and setrec2v 10008 using the old definition before I found the new one. I decided to change definitions for two reasons. First, as John Horton Conway noted in the Appendix to Part Zero of On Numbers and Games, mathematicians should not be caught up in any particular formalization, such as ZF set theory. Instead, they should work under whatever framework best suits the problem, and the formal bases used for different problems can be shown to be equivalent. Thus, Conway preferred defining surreal numbers as equivalence classes of surreal number forms, rather than sign-expansions. Although sign-expansions are easier to implement in ZF set theory, Conway argued that "formalisation within some particular axiomatic set theory is irrelevant". Furthermore, one of the most remarkable properties of the theory of surreal numbers is that it generates so much from almost nothing. Using sign-expansions as the formal definition destroys the beauty of surreal numbers, because ordinals are already built in. For this reason, I replaced the old definition of setrecs, which also relied heavily on ordinal numbers. On the other hand, both surreal numbers and the elementary definition of setrecs immediately generate the ordinal numbers from a (relatively) very simple set-theoretical basis. Second, although it is still complicated to formalize the theory of recursively generated sets within ZF set theory, it is actually simpler and more natural to do so with set theory directly than with the theory of ordinal numbers. As Conway wrote, indexing the "birthdays" of sets is and should be unnecessary. Using an elementary definition for setrecs removes the reliance on the previously developed theory of ordinal numbers, allowing proofs to be simpler and more direct. Formalizing surreal numbers within Metamath is probably still not in the spirit of Conway. He said that "attempts to force arbitrary theories into a single formal straitjacket... produce unnecessarily cumbrous and inelegant contortions." Nevertheless, Metamath has proven to be much more versatile than it seems at first, and I think the theory of surreal numbers can be natural while fitting well into the Metamath framework. The difficulty in writing a definition in Metamath for setrecs(𝐹) is that the necessary properties to prove are self-referential (see setrec1 10002 and setrec2v 10008), so we cannot simply write the properties we want inside a class abstraction as with most definitions. As noted in the comment of df-rdg 8449, this is not actually a requirement of the Metamath language, but we would like to be able to eliminate all definitions by direct mechanical substitution. We cannot define setrecs using a class abstraction directly, because nothing about its individual elements tells us whether they are in the set. We need to know about previous elements first. One way of getting around this problem without indexing is by defining setrecs(𝐹) as a union or intersection of suitable sets. Thus, instead of using a class abstraction for the elements of setrecs(𝐹), which seems to be impossible, we can use a class abstraction for supersets or subsets of setrecs(𝐹), which "know" about multiple individual elements at a time. Note that we cannot define setrecs(𝐹) as an intersection of sets, because in general it is a proper class, so any supersets would also be proper classes. However, a proper class can be a union of sets, as long as the collection of such sets is a proper class. Therefore, it is feasible to define setrecs(𝐹) as a union of a class abstraction. If setrecs(𝐹) = ∪ 𝐴, the elements of A must be subsets of setrecs(𝐹) which together include everything recursively generated by 𝐹. We can do this by letting 𝐴 be the class of sets 𝑥 whose elements are all recursively generated by 𝐹. One necessary condition is that each element of a given 𝑥 ∈ 𝐴 must be generated by 𝐹 when applied to a previous element 𝑦 ∈ 𝐴. In symbols, ∀𝑥 ∈ 𝐴∃𝑦 ∈ 𝐴(𝑦 ⊆ 𝑥 ∧ 𝑥 ⊆ (𝐹‘𝑦))}. However, this is not sufficient. All fixed points 𝑥 of 𝐹 will satisfy this condition whether they should be in setrecs(𝐹) or not. If we replace the subset relation with the proper subset relation, 𝑥 cannot be the empty set, even though the empty set should be in 𝐴. Therefore this condition cannot be used in the definition, even if we can find a way to avoid making it circular. A better strategy is to find a necessary and sufficient condition for all the elements of a set 𝑦 ∈ 𝐴 to be generated by 𝐹 when applied only to sets of previously generated elements within 𝑦. For example, taking 𝐹 to be the successor function, we can let 𝐴 = On rather than 𝒫 On, and we will still have ∪ 𝐴 = On as required. This gets rid of the circularity of the definition, since we should have a condition to test whether a given set 𝑦 is in 𝐴 without knowing about any of the other elements of 𝐴. The definition I ended up using accomplishes this using induction: 𝐴 is defined as the class of sets 𝑦 for which a sort of induction on the elements of 𝑦 holds. However, when creating a definition for setrecs that did not rely on ordinal numbers, I tried at first to write a definition using the well-founded relation predicate, Fr. I thought that this would be simple to do once I found a suitable definition using induction, just as the least- element principle is equivalent to induction on the positive integers. If we let 𝑅 = {⟨𝑎, 𝑏⟩ ∣ (𝐹‘𝑎) ⊆ 𝑏}, then (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥∀𝑧 ∈ 𝑥¬ (𝐹‘𝑧) ⊆ 𝑦)). On 22-Jul-2020 I came up with the following definition (Version 1) phrased in terms of induction: ∪ {𝑦 ∣ ∀𝑧 (∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ∈ 𝑧 → (𝐹‘𝑤) ∈ 𝑧)) → 𝑦 ∈ 𝑧)} In Aug-2020 I came up with an equivalent definition with the goal of phrasing it in terms of the relation Fr. It is the contrapositive of the previous one with 𝑧 replaced by its complement. ∪ {𝑦 ∣ ∀𝑧 (𝑦 ∈ 𝑧 → ∃𝑤(𝑤 ⊆ 𝑦 ∧ (𝐹‘𝑤) ∈ 𝑧 ∧ ¬ 𝑤 ∈ 𝑧))} These definitions didn't work because the induction didn't "get off the ground." If 𝑧 does not contain the empty set, the condition (∀𝑤...𝑦 ∈ 𝑧 fails, so 𝑦 = ∅ doesn't get included in 𝐴 even though it should. This could be fixed by adding the base case as a separate requirement, but the subtler problem would remain that rather than a set of "acceptable" sets, what we really need is a collection 𝑧 of all individuals that have been generated so far. So one approach is to replace every occurrence of ∈ 𝑧 with ⊆ 𝑧, making 𝑧 a set of individuals rather than a family of sets. That solves this problem, but it complicates the foundedness version of the definition, which looked cleaner in Version 1. There was another problem with Version 1. If we let 𝐹 be the power set function, then the induction in the inductive version works for 𝑧 being the class of transitive sets, restricted to subsets of 𝑦. Therefore, 𝑦 must be transitive by definition of 𝑧. This doesn't affect the union of all such 𝑦, but it may or may not be desirable. The problem is that 𝐹 is only applied to transitive sets, because of the strong requirement 𝑤 ∈ 𝑧, so the definition requires the additional constraint (𝑎 ⊆ 𝑏 → (𝐹‘𝑎) ⊆ (𝐹‘𝑏)) in order to work. This issue can also be avoided by replacing ∈ 𝑧 with ⊆ 𝑧. The induction version of the result is used in the final definition. Version 2: (18-Aug-2020) Induction: ∪ {𝑦 ∣ ∀𝑧 (∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)} Foundedness: ∪ {𝑦 ∣ ∀𝑧(𝑦 ∩ 𝑧 ≠ ∅ → ∃𝑤(𝑤 ⊆ 𝑦 ∧ 𝑤 ∩ 𝑧 = ∅ ∧ (𝐹‘𝑤) ∩ 𝑧 ≠ ∅))} In the induction version, not only does 𝑧 include all the elements of 𝑦, but it must include the elements of (𝐹‘𝑤) for 𝑤 ⊆ (𝑦 ∩ 𝑧) even if those elements of (𝐹‘𝑤) are not in 𝑦. We shouldn't care about any of the elements of 𝑧 outside 𝑦, but this detail doesn't affect the correctness of the definition. If we replaced (𝐹‘𝑤) in the definition by ((𝐹‘𝑤) ∩ 𝑦), we would get the same class for setrecs(𝐹). Suppose we could find a 𝑧 for which the condition fails for a given 𝑦 under the changed definition. Then the antecedent would be true, but 𝑦 ⊆ 𝑧 would be false. We could then simply add all elements of (𝐹‘𝑤) outside of 𝑦 for any 𝑤 ⊆ 𝑦, which we can do because all the classes involved are sets. This is not trivial and requires the axioms of union, power set, and replacement. However, the expanded 𝑧 fails the condition under the Metamath definition. The other direction is easier. If a certain 𝑧 fails the Metamath definition, then all (𝐹‘𝑤) ⊆ 𝑧 for 𝑤 ⊆ (𝑦 ∩ 𝑧), and in particular ((𝐹‘𝑤) ∩ 𝑦) ⊆ 𝑧. The foundedness version is starting to look more like ax-reg 9616! We want to take advantage of the preexisting relation Fr, which seems closely related to our foundedness definition. Since we only care about the elements of 𝑧 which are subsets of 𝑦, we can restrict 𝑧 to 𝑦 in the foundedness definition. Furthermore, instead of quantifying over 𝑤, quantify over the elements 𝑣 ∈ 𝑧 overlapping with 𝑤. Versions 3, 4, and 5 are all equivalent to Version 2. Version 3 - Foundedness (5-Sep-2020): ∪ {𝑦 ∣ ∀𝑧((𝑧 ⊆ 𝑦 ∧ 𝑧 ≠ ∅) → ∃𝑣 ∈ 𝑧∃𝑤(𝑤 ⊆ 𝑦 ∧ 𝑤 ∩ 𝑧 = ∅ ∧ 𝑣 ∈ (𝐹‘𝑤)))} Now, if we replace (𝐹‘𝑤) by ((𝐹‘𝑤) ∩ 𝑦), we do not change the definition. We already know that 𝑣 ∈ 𝑦 since 𝑣 ∈ 𝑧 and 𝑧 ⊆ 𝑦. All we need to show in order to prove that this change leads to an equivalent definition is to find To make our definition look exactly like df-fr 5609, we add another variable 𝑢 representing the nonexistent element of 𝑤 in 𝑧. Version 4 - Foundedness (6-Sep-2020): ∪ {𝑦 ∣ ∀𝑧((𝑧 ⊆ 𝑦 ∧ 𝑧 ≠ ∅) → ∃𝑣 ∈ 𝑧∃𝑤∀𝑢 ∈ 𝑧(𝑤 ⊆ 𝑦 ∧ ¬ 𝑢 ∈ 𝑤 ∧ 𝑣 ∈ (𝐹‘𝑤)) This is so close to df-fr 5609; the only change needed is to switch ∃𝑤 with ∀𝑢 ∈ 𝑧. Unfortunately, I couldn't find any way to switch the quantifiers without interfering with the definition. Maybe there is a definition equivalent to this one that uses Fr, but I couldn't find one. Yet, we can still find a remarkable similarity between Foundedness Version 2 and ax-reg 9616. Rather than a disjoint element of 𝑧, there's a disjoint coverer of an element of 𝑧. Finally, here's a different dead end I followed: To clean up our foundedness definition, we keep 𝑧 as a family of sets 𝑦 but allow 𝑤 to be any subset of ∪ 𝑧 in the induction. With this stronger induction, we can also allow for the stronger requirement 𝒫 𝑦 ⊆ 𝑧 rather than only 𝑦 ∈ 𝑧. This will help improve the foundedness version. Version 1.1 (28-Aug-2020) Induction: ∪ {𝑦 ∣ ∀𝑧(∀𝑤 (𝑤 ⊆ 𝑦 → (𝑤 ⊆ ∪ 𝑧 → (𝐹‘𝑤) ∈ 𝑧)) → 𝒫 𝑦 ⊆ 𝑧)} Foundedness: ∪ {𝑦 ∣ ∀𝑧(∃𝑎(𝑎 ⊆ 𝑦 ∧ 𝑎 ∈ 𝑧) → ∃𝑤(𝑤 ⊆ 𝑦 ∧ 𝑤 ∩ ∩ 𝑧 = ∅ ∧ (𝐹‘𝑤) ∈ 𝑧))} ( Edit (Aug 31) - this isn't true! Nothing forces the subset of an element of 𝑧 to be in 𝑧. Version 2 does not have this issue. ) Similarly, we could allow 𝑤 to be any subset of any element of 𝑧 rather than any subset of ∪ 𝑧. I think this has the same problem. We want to take advantage of the preexisting relation Fr, which seems closely related to our foundedness definition. Since we only care about the elements of 𝑧 which are subsets of 𝑦, we can restrict 𝑧 to 𝒫 𝑦 in the foundedness definition: Version 1.2 (31-Aug-2020) Foundedness: ∪ {𝑦 ∣ ∀𝑧((𝑧 ⊆ 𝒫 𝑦 ∧ 𝑧 ≠ ∅) → ∃𝑤(𝑤 ∈ 𝒫 𝑦 ∧ 𝑤 ∩ ∩ 𝑧 = ∅ ∧ (𝐹‘𝑤) ∈ 𝑧))} Now this looks more like df-fr 5609! The last step necessary to be able to use Fr directly in our definition is to replace (𝐹‘𝑤) with its own setvar variable, corresponding to 𝑦 in df-fr 5609. This definition is incorrect, though, since there's nothing forcing the subset of an element of 𝑧 to be in 𝑧. Version 1.3 (31-Aug-2020) Induction: ∪ {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ ∪ 𝑧 → (𝑤 ∈ 𝑧 ∧ (𝐹‘𝑤) ∈ 𝑧))) → 𝒫 𝑦 ⊆ 𝑧)} Foundedness: ∪ {𝑦 ∣ ∀𝑧((𝑧 ⊆ 𝒫 𝑦 ∧ 𝑧 ≠ ∅) → ∃𝑤(𝑤 ∈ 𝒫 𝑦 ∧ 𝑤 ∩ ∩ 𝑧 = ∅ ∧ (𝑤 ∈ 𝑧 ∨ (𝐹‘𝑤) ∈ 𝑧)))} 𝑧 must contain the supersets of each of its elements in the foundedness version, and we can't make any restrictions on 𝑧 or 𝐹, so this doesn't work. Let's try letting R be the covering relation 𝑅 = {⟨𝑎, 𝑏⟩ ∣ 𝑏 ∈ (𝐹‘𝑎)} to solve the transitivity issue (i.e. that if 𝐹 is the power set relation, 𝐴 consists only of transitive sets). The set (𝐹‘𝑤) corresponds to the variable 𝑦 in df-fr 5609. Thus, in our case, df-fr 5609 is equivalent to (𝑅 Fr 𝐴 ↔ ∀𝑧((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑤((𝐹‘𝑤) ∈ 𝑧 ∧ ¬ ∃𝑣 ∈ 𝑧𝑣𝑅(𝐹‘𝑤))). Substituting our relation 𝑅 gives (𝑅 Fr 𝐴 ↔ ∀𝑧((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑤((𝐹‘𝑤) ∈ 𝑧 ∧ ¬ ∃𝑣 ∈ 𝑧(𝐹‘𝑤) ∈ (𝐹‘𝑣))) This doesn't work for non-injective 𝐹 because we need all 𝑧 to be straddlers, but we don't necessarily need all-straddlers; loops within z are fine for non-injective F. Consider the foundedness form of Version 1. We want to show ¬ 𝑤 ∈ 𝑧 ↔ ∀𝑣 ∈ 𝑧¬ 𝑣𝑅(𝐹‘𝑤) so we can replace one with the other. Negate both sides: 𝑤 ∈ 𝑧 ↔ ∃𝑣 ∈ 𝑧𝑣𝑅(𝐹‘𝑤) If 𝐹 is injective, then we should be able to pick a suitable R, being careful about the above problem for some F (for example z = transitivity) when changing the antecedent y e. z' to z =/= (/). If we're clever, we can get rid of the injectivity requirement. The forward direction of the above equivalence always holds, but the key is that although the backwards direction doesn't hold in general, we can always find some z' where it doesn't work for 𝑤 itself. If there exists a z' where the version with the w condition fails, then there exists a z' where the version with the v condition also fails. However, Version 1 is not a correct definition, so this doesn't work either. (Contributed by Emmett Weisz, 18-Aug-2020.) (New usage is discouraged.)
⊢ setrecs(𝐹) = ∪ {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)}
Theorem	setrec1lem1 9996*	Lemma for setrec1 10002. This is a utility theorem showing the equivalence of the statement 𝑋 ∈ 𝑌 and its expanded form. The proof uses elabg 3633 and equivalence theorems. Variable 𝑌 is the class of sets 𝑦 that are recursively generated by the function 𝐹. In other words, 𝑦 ∈ 𝑌 iff by starting with the empty set and repeatedly applying 𝐹 to subsets 𝑤 of our set, we will eventually generate all the elements of 𝑌. In this theorem, 𝑋 is any element of 𝑌, and 𝑉 is any class. (Contributed by Emmett Weisz, 16-Oct-2020.) (New usage is discouraged.)
⊢ 𝑌 = {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)}    &   ⊢ (𝜑 → 𝑋 ∈ 𝑉)    ⇒   ⊢ (𝜑 → (𝑋 ∈ 𝑌 ↔ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑋 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑋 ⊆ 𝑧)))
Theorem	setrec1lem2 9997*	Lemma for setrec1 10002. If a family of sets are all recursively generated by 𝐹, so is their union. In this theorem, 𝑋 is a family of sets which are all elements of 𝑌, and 𝑉 is any class. Use dfss3 3902, equivalence and equality theorems, and unissb at the end. Sandwich with applications of setrec1lem1. (Contributed by Emmett Weisz, 24-Jan-2021.) (New usage is discouraged.)
⊢ 𝑌 = {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)}    &   ⊢ (𝜑 → 𝑋 ∈ 𝑉)    &   ⊢ (𝜑 → 𝑋 ⊆ 𝑌)    ⇒   ⊢ (𝜑 → ∪ 𝑋 ∈ 𝑌)
Theorem	bnd2d 9998*	Deduction form of bnd2 9986. (Contributed by Emmett Weisz, 19-Jan-2021.)
⊢ (𝜑 → 𝐴 ∈ V)    &   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜓)    ⇒   ⊢ (𝜑 → ∃𝑧(𝑧 ⊆ 𝐵 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑧 𝜓))
Theorem	setrec1lem3 9999*	Lemma for setrec1 10002. If each element 𝑎 of 𝐴 is covered by a set 𝑥 recursively generated by 𝐹, then there is a single such set covering all of 𝐴. The set is constructed explicitly using setrec1lem2 9997. It turns out that 𝑥 = 𝐴 also works, i.e., given the hypotheses it is possible to prove that 𝐴 ∈ 𝑌. I don't know if proving this fact directly using setrec1lem1 9996 would be any easier than the current proof using setrec1lem2 9997, and it would only slightly simplify the proof of setrec1 10002. Other than the use of bnd2d 9998, this is a purely technical theorem for rearranging notation from that of setrec1lem2 9997 to that of setrec1 10002. (Contributed by Emmett Weisz, 20-Jan-2021.) (New usage is discouraged.)
⊢ 𝑌 = {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)}    &   ⊢ (𝜑 → 𝐴 ∈ V)    &   ⊢ (𝜑 → ∀𝑎 ∈ 𝐴 ∃𝑥(𝑎 ∈ 𝑥 ∧ 𝑥 ∈ 𝑌))    ⇒   ⊢ (𝜑 → ∃𝑥(𝐴 ⊆ 𝑥 ∧ 𝑥 ∈ 𝑌))
Theorem	spcdvw 10000*	A version of spcdv 3552 where 𝜓 and 𝜒 are direct substitutions of each other. This theorem is useful because it does not require 𝜑 and 𝑥 to be distinct variables. (Contributed by Emmett Weisz, 12-Apr-2020.)
⊢ (𝜑 → 𝐴 ∈ 𝐵)    &   ⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜒))    ⇒   ⊢ (𝜑 → (∀𝑥𝜓 → 𝜒))