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Theorem List for Metamath Proof Explorer - 3601-3700   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremssrdv 3601* Deduction rule based on subclass definition. (Contributed by NM, 15-Nov-1995.)
(𝜑 → (𝑥𝐴𝑥𝐵))       (𝜑𝐴𝐵)

Theoremsstr2 3602 Transitivity of subclasses. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 24-Jun-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
(𝐴𝐵 → (𝐵𝐶𝐴𝐶))

Theoremsstr 3603 Transitivity of subclasses. Theorem 6 of [Suppes] p. 23. (Contributed by NM, 5-Sep-2003.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)

Theoremsstri 3604 Subclass transitivity inference. (Contributed by NM, 5-May-2000.)
𝐴𝐵    &   𝐵𝐶       𝐴𝐶

Theoremsstrd 3605 Subclass transitivity deduction. (Contributed by NM, 2-Jun-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)

Theoremsyl5ss 3606 Subclass transitivity deduction. (Contributed by NM, 6-Feb-2014.)
𝐴𝐵    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)

Theoremsyl6ss 3607 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
(𝜑𝐴𝐵)    &   𝐵𝐶       (𝜑𝐴𝐶)

Theoremsylan9ss 3608 A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
(𝜑𝐴𝐵)    &   (𝜓𝐵𝐶)       ((𝜑𝜓) → 𝐴𝐶)

Theoremsylan9ssr 3609 A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.)
(𝜑𝐴𝐵)    &   (𝜓𝐵𝐶)       ((𝜓𝜑) → 𝐴𝐶)

Theoremeqss 3610 The subclass relationship is antisymmetric. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 21-May-1993.)
(𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))

Theoremeqssi 3611 Infer equality from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 9-Sep-1993.)
𝐴𝐵    &   𝐵𝐴       𝐴 = 𝐵

Theoremeqssd 3612 Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 27-Jun-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐵𝐴)       (𝜑𝐴 = 𝐵)

Theoremsssseq 3613 If a class is a subclass of another class, the classes are equal iff the other class is a subclass of the first class. (Contributed by AV, 23-Dec-2020.)
(𝐵𝐴 → (𝐴𝐵𝐴 = 𝐵))

Theoremeqrd 3614 Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.) (Proof shortened by BJ, 1-Dec-2021.)
𝑥𝜑    &   𝑥𝐴    &   𝑥𝐵    &   (𝜑 → (𝑥𝐴𝑥𝐵))       (𝜑𝐴 = 𝐵)

TheoremeqrdOLD 3615 Obsolete proof of eqrd 3614 as of 1-Dec-2021. (Contributed by Thierry Arnoux, 21-Mar-2017.) (Proof modification is discouraged.) (New usage is discouraged.)
𝑥𝜑    &   𝑥𝐴    &   𝑥𝐵    &   (𝜑 → (𝑥𝐴𝑥𝐵))       (𝜑𝐴 = 𝐵)

Theoremssid 3616 Any class is a subclass of itself. Exercise 10 of [TakeutiZaring] p. 18. (Contributed by NM, 21-Jun-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
𝐴𝐴

Theoremssv 3617 Any class is a subclass of the universal class. (Contributed by NM, 31-Oct-1995.)
𝐴 ⊆ V

Theoremsseq1 3618 Equality theorem for subclasses. (Contributed by NM, 24-Jun-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
(𝐴 = 𝐵 → (𝐴𝐶𝐵𝐶))

Theoremsseq2 3619 Equality theorem for the subclass relationship. (Contributed by NM, 25-Jun-1998.)
(𝐴 = 𝐵 → (𝐶𝐴𝐶𝐵))

Theoremsseq12 3620 Equality theorem for the subclass relationship. (Contributed by NM, 31-May-1999.)
((𝐴 = 𝐵𝐶 = 𝐷) → (𝐴𝐶𝐵𝐷))

Theoremsseq1i 3621 An equality inference for the subclass relationship. (Contributed by NM, 18-Aug-1993.)
𝐴 = 𝐵       (𝐴𝐶𝐵𝐶)

Theoremsseq2i 3622 An equality inference for the subclass relationship. (Contributed by NM, 30-Aug-1993.)
𝐴 = 𝐵       (𝐶𝐴𝐶𝐵)

Theoremsseq12i 3623 An equality inference for the subclass relationship. (Contributed by NM, 31-May-1999.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
𝐴 = 𝐵    &   𝐶 = 𝐷       (𝐴𝐶𝐵𝐷)

Theoremsseq1d 3624 An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
(𝜑𝐴 = 𝐵)       (𝜑 → (𝐴𝐶𝐵𝐶))

Theoremsseq2d 3625 An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
(𝜑𝐴 = 𝐵)       (𝜑 → (𝐶𝐴𝐶𝐵))

Theoremsseq12d 3626 An equality deduction for the subclass relationship. (Contributed by NM, 31-May-1999.)
(𝜑𝐴 = 𝐵)    &   (𝜑𝐶 = 𝐷)       (𝜑 → (𝐴𝐶𝐵𝐷))

Theoremeqsstri 3627 Substitution of equality into a subclass relationship. (Contributed by NM, 16-Jul-1995.)
𝐴 = 𝐵    &   𝐵𝐶       𝐴𝐶

Theoremeqsstr3i 3628 Substitution of equality into a subclass relationship. (Contributed by NM, 19-Oct-1999.)
𝐵 = 𝐴    &   𝐵𝐶       𝐴𝐶

Theoremsseqtri 3629 Substitution of equality into a subclass relationship. (Contributed by NM, 28-Jul-1995.)
𝐴𝐵    &   𝐵 = 𝐶       𝐴𝐶

Theoremsseqtr4i 3630 Substitution of equality into a subclass relationship. (Contributed by NM, 4-Apr-1995.)
𝐴𝐵    &   𝐶 = 𝐵       𝐴𝐶

Theoremeqsstrd 3631 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴 = 𝐵)    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)

Theoremeqsstr3d 3632 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐵 = 𝐴)    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)

Theoremsseqtrd 3633 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐵 = 𝐶)       (𝜑𝐴𝐶)

Theoremsseqtr4d 3634 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐶 = 𝐵)       (𝜑𝐴𝐶)

Theorem3sstr3i 3635 Substitution of equality in both sides of a subclass relationship. (Contributed by NM, 13-Jan-1996.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
𝐴𝐵    &   𝐴 = 𝐶    &   𝐵 = 𝐷       𝐶𝐷

Theorem3sstr4i 3636 Substitution of equality in both sides of a subclass relationship. (Contributed by NM, 13-Jan-1996.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
𝐴𝐵    &   𝐶 = 𝐴    &   𝐷 = 𝐵       𝐶𝐷

Theorem3sstr3g 3637 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
(𝜑𝐴𝐵)    &   𝐴 = 𝐶    &   𝐵 = 𝐷       (𝜑𝐶𝐷)

Theorem3sstr4g 3638 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 16-Aug-1994.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
(𝜑𝐴𝐵)    &   𝐶 = 𝐴    &   𝐷 = 𝐵       (𝜑𝐶𝐷)

Theorem3sstr3d 3639 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
(𝜑𝐴𝐵)    &   (𝜑𝐴 = 𝐶)    &   (𝜑𝐵 = 𝐷)       (𝜑𝐶𝐷)

Theorem3sstr4d 3640 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 30-Nov-1995.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
(𝜑𝐴𝐵)    &   (𝜑𝐶 = 𝐴)    &   (𝜑𝐷 = 𝐵)       (𝜑𝐶𝐷)

Theoremsyl5eqss 3641 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
𝐴 = 𝐵    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)

Theoremsyl5eqssr 3642 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
𝐵 = 𝐴    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)

Theoremsyl6sseq 3643 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   𝐵 = 𝐶       (𝜑𝐴𝐶)

Theoremsyl6sseqr 3644 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   𝐶 = 𝐵       (𝜑𝐴𝐶)

Theoremsyl5sseq 3645 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
𝐵𝐴    &   (𝜑𝐴 = 𝐶)       (𝜑𝐵𝐶)

Theoremsyl5sseqr 3646 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
𝐵𝐴    &   (𝜑𝐶 = 𝐴)       (𝜑𝐵𝐶)

Theoremsyl6eqss 3647 A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
(𝜑𝐴 = 𝐵)    &   𝐵𝐶       (𝜑𝐴𝐶)

Theoremsyl6eqssr 3648 A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
(𝜑𝐵 = 𝐴)    &   𝐵𝐶       (𝜑𝐴𝐶)

Theoremeqimss 3649 Equality implies the subclass relation. (Contributed by NM, 21-Jun-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
(𝐴 = 𝐵𝐴𝐵)

Theoremeqimss2 3650 Equality implies the subclass relation. (Contributed by NM, 23-Nov-2003.)
(𝐵 = 𝐴𝐴𝐵)

Theoremeqimssi 3651 Infer subclass relationship from equality. (Contributed by NM, 6-Jan-2007.)
𝐴 = 𝐵       𝐴𝐵

Theoremeqimss2i 3652 Infer subclass relationship from equality. (Contributed by NM, 7-Jan-2007.)
𝐴 = 𝐵       𝐵𝐴

Theoremnssne1 3653 Two classes are different if they don't include the same class. (Contributed by NM, 23-Apr-2015.)
((𝐴𝐵 ∧ ¬ 𝐴𝐶) → 𝐵𝐶)

Theoremnssne2 3654 Two classes are different if they are not subclasses of the same class. (Contributed by NM, 23-Apr-2015.)
((𝐴𝐶 ∧ ¬ 𝐵𝐶) → 𝐴𝐵)

Theoremnss 3655* Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
𝐴𝐵 ↔ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))

Theoremnelss 3656 Demonstrate by witnesses that two classes lack a subclass relation. (Contributed by Stefan O'Rear, 5-Feb-2015.)
((𝐴𝐵 ∧ ¬ 𝐴𝐶) → ¬ 𝐵𝐶)

Theoremssrexf 3657 restricted existential quantification follows from a subclass relationship. (Contributed by Glauco Siliprandi, 20-Apr-2017.)
𝑥𝐴    &   𝑥𝐵       (𝐴𝐵 → (∃𝑥𝐴 𝜑 → ∃𝑥𝐵 𝜑))

Theoremssralv 3658* Quantification restricted to a subclass. (Contributed by NM, 11-Mar-2006.)
(𝐴𝐵 → (∀𝑥𝐵 𝜑 → ∀𝑥𝐴 𝜑))

Theoremssrexv 3659* Existential quantification restricted to a subclass. (Contributed by NM, 11-Jan-2007.)
(𝐴𝐵 → (∃𝑥𝐴 𝜑 → ∃𝑥𝐵 𝜑))

Theoremralss 3660* Restricted universal quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.)
(𝐴𝐵 → (∀𝑥𝐴 𝜑 ↔ ∀𝑥𝐵 (𝑥𝐴𝜑)))

Theoremrexss 3661* Restricted existential quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.)
(𝐴𝐵 → (∃𝑥𝐴 𝜑 ↔ ∃𝑥𝐵 (𝑥𝐴𝜑)))

Theoremss2ab 3662 Class abstractions in a subclass relationship. (Contributed by NM, 3-Jul-1994.)
({𝑥𝜑} ⊆ {𝑥𝜓} ↔ ∀𝑥(𝜑𝜓))

Theoremabss 3663* Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))

Theoremssab 3664* Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)
(𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))

Theoremssabral 3665* The relation for a subclass of a class abstraction is equivalent to restricted quantification. (Contributed by NM, 6-Sep-2006.)
(𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥𝐴 𝜑)

Theoremss2abi 3666 Inference of abstraction subclass from implication. (Contributed by NM, 31-Mar-1995.)
(𝜑𝜓)       {𝑥𝜑} ⊆ {𝑥𝜓}

Theoremss2abdv 3667* Deduction of abstraction subclass from implication. (Contributed by NM, 29-Jul-2011.)
(𝜑 → (𝜓𝜒))       (𝜑 → {𝑥𝜓} ⊆ {𝑥𝜒})

Theoremabssdv 3668* Deduction of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006.)
(𝜑 → (𝜓𝑥𝐴))       (𝜑 → {𝑥𝜓} ⊆ 𝐴)

Theoremabssi 3669* Inference of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006.)
(𝜑𝑥𝐴)       {𝑥𝜑} ⊆ 𝐴

Theoremss2rab 3670 Restricted abstraction classes in a subclass relationship. (Contributed by NM, 30-May-1999.)
({𝑥𝐴𝜑} ⊆ {𝑥𝐴𝜓} ↔ ∀𝑥𝐴 (𝜑𝜓))

Theoremrabss 3671* Restricted class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
({𝑥𝐴𝜑} ⊆ 𝐵 ↔ ∀𝑥𝐴 (𝜑𝑥𝐵))

Theoremssrab 3672* Subclass of a restricted class abstraction. (Contributed by NM, 16-Aug-2006.)
(𝐵 ⊆ {𝑥𝐴𝜑} ↔ (𝐵𝐴 ∧ ∀𝑥𝐵 𝜑))

Theoremssrabdv 3673* Subclass of a restricted class abstraction (deduction rule). (Contributed by NM, 31-Aug-2006.)
(𝜑𝐵𝐴)    &   ((𝜑𝑥𝐵) → 𝜓)       (𝜑𝐵 ⊆ {𝑥𝐴𝜓})

Theoremrabssdv 3674* Subclass of a restricted class abstraction (deduction rule). (Contributed by NM, 2-Feb-2015.)
((𝜑𝑥𝐴𝜓) → 𝑥𝐵)       (𝜑 → {𝑥𝐴𝜓} ⊆ 𝐵)

Theoremss2rabdv 3675* Deduction of restricted abstraction subclass from implication. (Contributed by NM, 30-May-2006.)
((𝜑𝑥𝐴) → (𝜓𝜒))       (𝜑 → {𝑥𝐴𝜓} ⊆ {𝑥𝐴𝜒})

Theoremss2rabi 3676 Inference of restricted abstraction subclass from implication. (Contributed by NM, 14-Oct-1999.)
(𝑥𝐴 → (𝜑𝜓))       {𝑥𝐴𝜑} ⊆ {𝑥𝐴𝜓}

Theoremrabss2 3677* Subclass law for restricted abstraction. (Contributed by NM, 18-Dec-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝐴𝐵 → {𝑥𝐴𝜑} ⊆ {𝑥𝐵𝜑})

Theoremssab2 3678* Subclass relation for the restriction of a class abstraction. (Contributed by NM, 31-Mar-1995.)
{𝑥 ∣ (𝑥𝐴𝜑)} ⊆ 𝐴

Theoremssrab2 3679* Subclass relation for a restricted class. (Contributed by NM, 19-Mar-1997.)
{𝑥𝐴𝜑} ⊆ 𝐴

Theoremssrab3 3680* Subclass relation for a restricted class abstraction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
𝐵 = {𝑥𝐴𝜑}       𝐵𝐴

Theoremssrabeq 3681* If the restricting class of a restricted class abstraction is a subset of this restricted class abstraction, it is equal to this restricted class abstraction. (Contributed by Alexander van der Vekens, 31-Dec-2017.)
(𝑉 ⊆ {𝑥𝑉𝜑} ↔ 𝑉 = {𝑥𝑉𝜑})

Theoremrabssab 3682 A restricted class is a subclass of the corresponding unrestricted class. (Contributed by Mario Carneiro, 23-Dec-2016.)
{𝑥𝐴𝜑} ⊆ {𝑥𝜑}

Theoremuniiunlem 3683* A subset relationship useful for converting union to indexed union using dfiun2 4545 or dfiun2g 4543 and intersection to indexed intersection using dfiin2 4546. (Contributed by NM, 5-Oct-2006.) (Proof shortened by Mario Carneiro, 26-Sep-2015.)
(∀𝑥𝐴 𝐵𝐷 → (∀𝑥𝐴 𝐵𝐶 ↔ {𝑦 ∣ ∃𝑥𝐴 𝑦 = 𝐵} ⊆ 𝐶))

Theoremdfpss2 3684 Alternate definition of proper subclass. (Contributed by NM, 7-Feb-1996.)
(𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐴 = 𝐵))

Theoremdfpss3 3685 Alternate definition of proper subclass. (Contributed by NM, 7-Feb-1996.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝐴𝐵 ↔ (𝐴𝐵 ∧ ¬ 𝐵𝐴))

Theorempsseq1 3686 Equality theorem for proper subclass. (Contributed by NM, 7-Feb-1996.)
(𝐴 = 𝐵 → (𝐴𝐶𝐵𝐶))

Theorempsseq2 3687 Equality theorem for proper subclass. (Contributed by NM, 7-Feb-1996.)
(𝐴 = 𝐵 → (𝐶𝐴𝐶𝐵))

Theorempsseq1i 3688 An equality inference for the proper subclass relationship. (Contributed by NM, 9-Jun-2004.)
𝐴 = 𝐵       (𝐴𝐶𝐵𝐶)

Theorempsseq2i 3689 An equality inference for the proper subclass relationship. (Contributed by NM, 9-Jun-2004.)
𝐴 = 𝐵       (𝐶𝐴𝐶𝐵)

Theorempsseq12i 3690 An equality inference for the proper subclass relationship. (Contributed by NM, 9-Jun-2004.)
𝐴 = 𝐵    &   𝐶 = 𝐷       (𝐴𝐶𝐵𝐷)

Theorempsseq1d 3691 An equality deduction for the proper subclass relationship. (Contributed by NM, 9-Jun-2004.)
(𝜑𝐴 = 𝐵)       (𝜑 → (𝐴𝐶𝐵𝐶))

Theorempsseq2d 3692 An equality deduction for the proper subclass relationship. (Contributed by NM, 9-Jun-2004.)
(𝜑𝐴 = 𝐵)       (𝜑 → (𝐶𝐴𝐶𝐵))

Theorempsseq12d 3693 An equality deduction for the proper subclass relationship. (Contributed by NM, 9-Jun-2004.)
(𝜑𝐴 = 𝐵)    &   (𝜑𝐶 = 𝐷)       (𝜑 → (𝐴𝐶𝐵𝐷))

Theorempssss 3694 A proper subclass is a subclass. Theorem 10 of [Suppes] p. 23. (Contributed by NM, 7-Feb-1996.)
(𝐴𝐵𝐴𝐵)

Theorempssne 3695 Two classes in a proper subclass relationship are not equal. (Contributed by NM, 16-Feb-2015.)
(𝐴𝐵𝐴𝐵)

Theorempssssd 3696 Deduce subclass from proper subclass. (Contributed by NM, 29-Feb-1996.)
(𝜑𝐴𝐵)       (𝜑𝐴𝐵)

Theorempssned 3697 Proper subclasses are unequal. Deduction form of pssne 3695. (Contributed by David Moews, 1-May-2017.)
(𝜑𝐴𝐵)       (𝜑𝐴𝐵)

Theoremsspss 3698 Subclass in terms of proper subclass. (Contributed by NM, 25-Feb-1996.)
(𝐴𝐵 ↔ (𝐴𝐵𝐴 = 𝐵))

Theorempssirr 3699 Proper subclass is irreflexive. Theorem 7 of [Suppes] p. 23. (Contributed by NM, 7-Feb-1996.)
¬ 𝐴𝐴

Theorempssn2lp 3700 Proper subclass has no 2-cycle loops. Compare Theorem 8 of [Suppes] p. 23. (Contributed by NM, 7-Feb-1996.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
¬ (𝐴𝐵𝐵𝐴)

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