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Most recent proofs    These are the 100 (Unicode, GIF) or 1000 (Unicode, GIF) most recent proofs in the iset.mm database for the Intuitionistic Logic Explorer. The iset.mm database is maintained on GitHub with master (stable) and develop (development) versions. This page was created from the commit given on the MPE Most Recent Proofs page. The database from that commit is also available here: iset.mm.

See the MPE Most Recent Proofs page for news and some useful links.

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Last updated on 9-Sep-2026 at 7:09 AM ET.
Recent Additions to the Intuitionistic Logic Explorer
DateLabelDescription
Theorem
 
27-Aug-2026prmdcz 12925 Primality is decidable. (Contributed by Jim Kingdon, 27-Aug-2026.)
(𝑁 ∈ ℤ → DECID 𝑁 ∈ ℙ)
 
27-Aug-2026zmincl 12020 The minumum of two integers is an integer. (Contributed by Jim Kingdon, 27-Aug-2026.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → inf({𝐴, 𝐵}, ℝ, < ) ∈ ℤ)
 
25-Aug-2026nn0sqdcq 13004 A nonnegative integer is a perfect square or not. This is similar to nn0sqdc 11160 but expresses the idea of being a perfect square as having a rational number which, when squared, gives the original number. (Contributed by Jim Kingdon, 25-Aug-2026.)
(𝑁 ∈ ℕ0 → DECID ∃𝑞 ∈ ℚ 𝑁 = (𝑞↑2))
 
25-Aug-2026qabscl 11857 The absolute value of a rational number is a rational number. (Contributed by Jim Kingdon, 25-Aug-2026.)
(𝐴 ∈ ℚ → (abs‘𝐴) ∈ ℚ)
 
25-Aug-2026nn0sqdc 11160 A nonnegative integer is a perfect square or not. (Contributed by Jim Kingdon, 25-Aug-2026.)
(𝑁 ∈ ℕ0 → DECID ∃𝑞 ∈ ℕ0 𝑁 = (𝑞↑2))
 
24-Aug-2026sqrtrirr 13005 The square root of a nonnegative integer is either rational or irrational. (Contributed by Jim Kingdon, 24-Aug-2026.)
(𝐴 ∈ ℕ0 → ((√‘𝐴) ∈ ℚ ∨ ((√‘𝐴) ∈ ℝ ∧ ∀𝑞 ∈ ℚ (√‘𝐴) # 𝑞)))
 
21-Aug-2026prmefexple 16227 Convert a bound on a power of a prime to a bound on the exponent. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Jim Kingdon, 21-Aug-2026.)
((𝐴 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ 𝐵 ∈ ℕ) → ((𝐴↑𝑁) ≤ 𝐵 ↔ 𝑁 ≤ (⌊‘((log‘𝐵) / (log‘𝐴)))))
 
20-Aug-2026zprmlogbap 16137 The logarithm of a natural number to a prime base is either rational or irrational.

The proof decomposes 𝑋 into 𝑚 ∈ ℕ and 𝑎 ∈ ℕ0 such that 𝑋 = ((𝐵↑𝑎) · 𝑚) (using nnmaxpw 12969). If 𝑚 = 1 the logarithm is 𝑎, which is rational. If 1 < 𝑚 then we can apply logbgcd1irrap 16125 to show that the logarithm is irrational. (Contributed by Jim Kingdon and Taylor Barrella, 20-Aug-2026.)

((𝑋 ∈ ℕ ∧ 𝐵 ∈ ℙ) → ((𝐵 logb 𝑋) ∈ ℚ ∨ ((𝐵 logb 𝑋) ∈ ℝ ∧ ∀𝑞 ∈ ℚ (𝐵 logb 𝑋) # 𝑞)))
 
20-Aug-2026zprmlogbaplem3 16136 Lemma for zprmlogbap 16137. Decomposing a natural number into a power of a prime base and a factor not divisible by that prime. (Contributed by Jim Kingdon, 20-Aug-2026.)
𝐽 = {𝑧 ∈ ℕ ∣ ¬ 𝐵 ∥ 𝑧}    &   𝐹 = (𝑥 ∈ 𝐽, 𝑦 ∈ ℕ0 ↦ ((𝐵↑𝑦) · 𝑥))    ⇒   ((𝑋 ∈ ℕ ∧ 𝐵 ∈ ℙ) → ∃𝑚 ∈ ℕ ∃𝑎 ∈ ℕ0 (¬ 𝐵 ∥ 𝑚 ∧ 𝑋 = ((𝐵↑𝑎) · 𝑚)))
 
20-Aug-2026zprmlogbaplem2 16135 Lemma for zprmlogbap 16137. The logarithm is either rational or irrational. (Contributed by Jim Kingdon, 20-Aug-2026.)
(𝜑 → 𝐵 ∈ ℙ)    &   (𝜑 → 𝑀 ∈ ℕ)    &   (𝜑 → ¬ 𝐵 ∥ 𝑀)    &   (𝜑 → 𝐴 ∈ ℕ0)    &   𝑋 = ((𝐵↑𝐴) · 𝑀)    ⇒   (𝜑 → ((𝐵 logb 𝑋) ∈ ℚ ∨ ((𝐵 logb 𝑋) ∈ ℝ ∧ ∀𝑞 ∈ ℚ (𝐵 logb 𝑋) # 𝑞)))
 
20-Aug-2026zprmlogbaplem1 16134 Lemma for zprmlogbap 16137. Rearranging an expression involving logarithms. (Contributed by Jim Kingdon, 20-Aug-2026.)
(𝜑 → 𝐵 ∈ ℙ)    &   (𝜑 → 𝑀 ∈ ℕ)    &   (𝜑 → ¬ 𝐵 ∥ 𝑀)    &   (𝜑 → 𝐴 ∈ ℕ0)    ⇒   (𝜑 → (𝐵 logb ((𝐵↑𝐴) · 𝑀)) = (𝐴 + (𝐵 logb 𝑀)))
 
20-Aug-2026flaplelt 10723 A basic property of the floor (greatest integer) function. (Contributed by Jim Kingdon, 20-Aug-2026.)
((𝐴 ∈ ℚ ∨ (𝐴 ∈ ℝ ∧ ∀𝑞 ∈ ℚ 𝐴 # 𝑞)) → ((⌊‘𝐴) ≤ 𝐴 ∧ 𝐴 < ((⌊‘𝐴) + 1)))
 
20-Aug-2026flapcl 10721 The floor (greatest integer) function yields an integer when applied to a number which is either rational or irrational. (Contributed by Jim Kingdon, 20-Aug-2026.)
((𝐴 ∈ ℚ ∨ (𝐴 ∈ ℝ ∧ ∀𝑞 ∈ ℚ 𝐴 # 𝑞)) → (⌊‘𝐴) ∈ ℤ)
 
20-Aug-2026irraddap 10056 The sum of an irrational number and a rational number is irrational. (Contributed by Jim Kingdon, 20-Aug-2026.)
(((𝐴 ∈ ℝ ∧ ∀𝑞 ∈ ℚ 𝐴 # 𝑞) ∧ 𝐵 ∈ ℚ) → ((𝐴 + 𝐵) ∈ ℝ ∧ ∀𝑞 ∈ ℚ (𝐴 + 𝐵) # 𝑞))
 
19-Aug-2026nnmaxpw 12969 The function 𝐹 that decomposes a number into its "odd" and "even" parts, which is to say the largest power of a base and largest divisor of the number not divisible by that base, is a bijection from pairs of a nonnegative integer and a number not divisible by that base to positive integers. (Contributed by Thierry Arnoux, 15-Aug-2017.) (Revised by Jim Kingdon, 19-Aug-2026.)
𝐽 = {𝑧 ∈ ℕ ∣ ¬ 𝐵 ∥ 𝑧}    &   𝐹 = (𝑥 ∈ 𝐽, 𝑦 ∈ ℕ0 ↦ ((𝐵↑𝑦) · 𝑥))    ⇒   (𝐵 ∈ (ℤ≥‘2) → 𝐹:(𝐽 × ℕ0)–1-1-onto→ℕ)
 
19-Aug-2026nnmaxpwlemparts 12968 Lemma for nnmaxpw 12969. Decomposing a number into parts. (Contributed by Jim Kingdon, 16-Nov-2021.) (Revised by Jim Kingdon, 19-Aug-2026.)
(𝐵 ∈ (ℤ≥‘2) → ((((𝑋 ∈ ℕ ∧ ¬ 𝐵 ∥ 𝑋) ∧ 𝑌 ∈ ℕ0) ∧ 𝐴 = ((𝐵↑𝑌) · 𝑋)) ↔ (𝐴 ∈ ℕ ∧ (𝑋 = (𝐴 / (𝐵↑(℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴)))) ∧ 𝑌 = (℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴))))))
 
19-Aug-2026nnmaxpwlemnfac 12967 Lemma for nnmaxpw 12969. Removing the powers of a base from a natural number produces a number not divisible by that base. (Contributed by Jim Kingdon, 16-Nov-2021.) (Revised by Jim Kingdon, 19-Aug-2026.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ (ℤ≥‘2)) → ¬ 𝐵 ∥ (𝐴 / (𝐵↑(℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴)))))
 
19-Aug-2026nnmaxpwlemndvds 12966 Lemma for nnmaxpw 12969. A natural number is not divisible by one more than the highest power of a base which divides it. (Contributed by Jim Kingdon, 17-Nov-2021.) (Revised by Jim Kingdon, 19-Aug-2026.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ (ℤ≥‘2)) → ¬ (𝐵↑((℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴)) + 1)) ∥ 𝐴)
 
19-Aug-2026nnmaxpwlemdvds 12965 Lemma for nnmaxpw 12969. A natural number is divisible by the highest power of a base which divides it. (Contributed by Jim Kingdon, 17-Nov-2021.) (Revised by Jim Kingdon, 19-Aug-2026.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ (ℤ≥‘2)) → (𝐵↑(℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴))) ∥ 𝐴)
 
18-Aug-2026nnmaxpwlemxy 12964 Lemma for nnmaxpw 12969. Another way of stating that decomposing a natural number into a power of a base and a number not divisible by that base is unique. (Contributed by Jim Kingdon, 16-Nov-2021.) (Revised by Jim Kingdon, 18-Aug-2026.)
(𝜑 → 𝑋 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ (ℤ≥‘2))    &   (𝜑 → ¬ 𝐵 ∥ 𝑋)    &   (𝜑 → 𝑌 ∈ ℕ0)    &   (𝜑 → 𝐴 = ((𝐵↑𝑌) · 𝑋))    ⇒   (𝜑 → (𝑋 = (𝐴 / (𝐵↑(℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴)))) ∧ 𝑌 = (℩𝑧 ∈ ℕ0 ((𝐵↑𝑧) ∥ 𝐴 ∧ ¬ (𝐵↑(𝑧 + 1)) ∥ 𝐴))))
 
18-Aug-2026pwbdvdseu 12963 A natural number has a unique highest power of a base which divides it. (Contributed by Jim Kingdon, 16-Nov-2021.) (Revised by Jim Kingdon, 18-Aug-2026.)
((𝑁 ∈ ℕ ∧ 𝐵 ∈ (ℤ≥‘2)) → ∃!𝑚 ∈ ℕ0 ((𝐵↑𝑚) ∥ 𝑁 ∧ ¬ (𝐵↑(𝑚 + 1)) ∥ 𝑁))
 
18-Aug-2026pwbdvdseulemle 12962 Lemma for pwbdvdseu 12963. Powers of a base which do and do not divide a natural number. (Contributed by Jim Kingdon, 17-Nov-2021.) (Revised by Jim Kingdon, 18-Aug-2026.)
(𝜑 → 𝑁 ∈ ℕ)    &   (𝜑 → 𝐴 ∈ ℕ0)    &   (𝜑 → 𝐵 ∈ ℕ0)    &   (𝜑 → 𝑃 ∈ ℕ)    &   (𝜑 → (𝑃↑𝐴) ∥ 𝑁)    &   (𝜑 → ¬ (𝑃↑(𝐵 + 1)) ∥ 𝑁)    ⇒   (𝜑 → 𝐴 ≤ 𝐵)
 
18-Aug-2026pwbdvds 12961 A natural number has a highest power of a base which divides it. (Contributed by Jim Kingdon, 16-Nov-2021.) (Revised by Jim Kingdon, 18-Aug-2026.)
((𝑁 ∈ ℕ ∧ 𝐵 ∈ (ℤ≥‘2)) → ∃𝑚 ∈ ℕ0 ((𝐵↑𝑚) ∥ 𝑁 ∧ ¬ (𝐵↑(𝑚 + 1)) ∥ 𝑁))
 
17-Aug-2026pwbdvdslemn 12960 Lemma for pwbdvds 12961. If a natural number has some power of a base which does not divide it, there is a highest power of the base which does divide it. (Contributed by Jim Kingdon, 14-Nov-2021.) (Revised by Jim Kingdon, 17-Aug-2026.)
(𝜑 → 𝑁 ∈ ℕ)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → ¬ (𝐵↑𝐴) ∥ 𝑁)    ⇒   (𝜑 → ∃𝑚 ∈ ℕ0 ((𝐵↑𝑚) ∥ 𝑁 ∧ ¬ (𝐵↑(𝑚 + 1)) ∥ 𝑁))
 
14-Aug-2026reaplog 16019 Apartness and the real natural logarithm. (Contributed by Jim Kingdon, 14-Aug-2026.)
((𝐴 ∈ ℝ+ ∧ 𝐵 ∈ ℝ+) → (𝐴 # 𝐵 ↔ (log‘𝐴) # (log‘𝐵)))
 
13-Aug-2026efap1p 15929 If the exponential of a number is apart from one plus that number, the number is apart from zero. To some extent can be thought of as the converse of efgt1p 12479. (Contributed by Jim Kingdon, 13-Aug-2026.)
((𝐴 ∈ ℝ ∧ (1 + 𝐴) # (exp‘𝐴)) → 𝐴 # 0)
 
6-Aug-2026relndmfv 5728 The value of a relation outside its domain is the empty set. (Contributed by Jim Kingdon, 6-Aug-2026.)
((Rel 𝐹 ∧ ¬ 𝐴 ∈ dom 𝐹) → (𝐹‘𝐴) = ∅)
 
1-Aug-2026wexmiddifxy 17165 Being able to subtract an arbitrary finite set from a finite set and get a finite set is equivalent to weak excluded middle. By adding additional conditions we can get a theorem which does not need weak excluded middle, at diffifi 7198. (Contributed by Jim Kingdon, 1-Aug-2026.)
(WEXMID ↔ ∀𝑥∀𝑦((𝑥 ∈ Fin ∧ 𝑦 ∈ Fin) → (𝑥 ∖ 𝑦) ∈ Fin))
 
1-Aug-2026wexmiddifxylem 17164 Lemma for wexmiddifxylem 17164. Showing weak excluded middle given a suitable finite set. (Contributed by Jim Kingdon, 1-Aug-2026.)
(({1o} ∖ {{𝑥 ∈ 1o ∣ 𝜑}}) ∈ Fin → DECID ¬ 𝜑)
 
31-Jul-2026rabid1o 17153 Converting between propositions and corresponding subsets of a singleton. (Contributed by Jim Kingdon, 31-Jul-2026.)
({𝑥 ∈ 1o ∣ 𝜑} = 1o ↔ 𝜑)
 
30-Jul-2026wexmiddc 17161 Weak excluded middle expressed using WEXMID implies decidability of a negated proposition. (Contributed by Jim Kingdon, 30-Jul-2026.)
(WEXMID → DECID ¬ 𝜑)
 
30-Jul-2026df-wexmid 17160 Weak excluded middle is the principle that any negated proposition is decidable. (Contributed by Jim Kingdon, 30-Jul-2026.)
(WEXMID ↔ ∀𝑝 ∈ 𝒫 1o(¬ 𝑝 = 1o ∨ ¬ ¬ 𝑝 = 1o))
 
29-Jul-2026wexmiddiffi 17163 Being able to subtract an arbitrary set from a finite set and get a finite set is equivalent to weak excluded middle. By adding additional conditions we can get a theorem which does not need weak excluded middle, at diffifi 7198. (Contributed by Jim Kingdon, 29-Jul-2026.)
(WEXMID ↔ ∀𝑥∀𝑦(𝑥 ∈ Fin → (𝑥 ∖ 𝑦) ∈ Fin))
 
29-Jul-2026wexmiddiffilem 17162 Lemma for wexmiddiffi 17163. The reverse direction, using different notation. (Contributed by Jim Kingdon, 29-Jul-2026.)
(∀𝑥∀𝑦(𝑥 ∈ Fin → (𝑥 ∖ 𝑦) ∈ Fin) → (¬ 𝜑 ∨ ¬ ¬ 𝜑))
 
24-Jul-2026stnot 17158 A proposition is double negation stable if and only if it is equivalent to a negated proposition. Here by "proposition" we mean a subset of a singleton (which is a choice which allows us to quantify over them). Posed as an exercise online by Yannick Forster. (Contributed by Jim Kingdon, 24-Jul-2026.)
(𝐴 ∈ 𝒫 1o → ((¬ ¬ 𝐴 = 1o → 𝐴 = 1o) ↔ ∃𝑦 ∈ 𝒫 1o(𝐴 = 1o ↔ ¬ 𝑦 = 1o)))
 
24-Jul-2026slotm 13464 A structure with an inhabited slot is inhabited. (Contributed by Jim Kingdon, 24-Jul-2026.)
(𝐸 = Slot (𝐸‘ndx) ∧ (𝐸‘ndx) ∈ ℕ)    ⇒   (𝐴 ∈ (𝐸‘𝐺) → ∃𝑗 𝑗 ∈ 𝐺)
 
22-Jul-2026alseu-no-surprise 17298 Demonstrate that there is never a "surprise" when using the "all some one" quantifier, that is, it is never possible for the consequent to be both always true and always false. This follows from als-no-surprise 17266 by alseuals 17284. See als-no-surprise 17266 for why ordinary "for all" with implication has no such property. (Contributed by David A. Wheeler, 22-Jul-2026.)
¬ (∀∃!𝑥(𝜑 → 𝜓) ∧ ∀∃!𝑥(𝜑 → ¬ 𝜓))
 
22-Jul-2026alseueu 17297 "The 𝜑 is 𝜓 " implies that exactly one thing is both 𝜑 and 𝜓. This is the half of dfalseu2 17296 that drops the universal conjunct; it does not reverse, so ∃!𝑥(𝜑 ∧ 𝜓) cannot be used in place of an "all some one" statement. (Contributed by David A. Wheeler, 22-Jul-2026.)
(∀∃!𝑥(𝜑 → 𝜓) → ∃!𝑥(𝜑 ∧ 𝜓))
 
22-Jul-2026dfalseu2 17296 An "all some one" statement is equivalent to its universal part conjoined with the claim that exactly one 𝑥 satisfies both 𝜑 and 𝜓. In other words, given ∀𝑥(𝜑 → 𝜓), requiring exactly one 𝑥 to satisfy 𝜑, which is what df-alseu 17281 requires, and requiring exactly one 𝑥 to satisfy (𝜑 ∧ 𝜓) come to the same thing. Read 𝜑 as "is a king" and 𝜓 as "is hungry": if every king is hungry, then "there is exactly one king" and "there is exactly one hungry king" say the same thing, so either of them, together with "every king is hungry", gives "the king is hungry".

The universal conjunct is what makes that work, and it cannot be dropped. ∃!𝑥(𝜑 ∧ 𝜓) on its own is strictly weaker than ∀∃!𝑥(𝜑 → 𝜓), since it is satisfied when many things are 𝜑 and just one of those is 𝜓, as in a region with five kings exactly one of whom is hungry; see alseueu 17297 for the one direction that does hold without it. Uniqueness attaches to the antecedent, not to the conjunction. Russell's analysis of a definite description is built the same way: its uniqueness clause constrains the description predicate alone, while the predication is a separate conjunct. See his worked example of "the father of Charles II was executed", [Russell1905] p. 482. (Contributed by David A. Wheeler, 22-Jul-2026.)

(∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥(𝜑 ∧ 𝜓)))
 
22-Jul-2026nfralseu 17295 Bound-variable hypothesis builder for "all some one" restricted to a class. This is the "all some one" counterpart of nfrals 17264. (Contributed by David A. Wheeler, 22-Jul-2026.)
Ⅎ𝑥𝐴    &   Ⅎ𝑥𝜑    &   Ⅎ𝑥𝜓    ⇒   Ⅎ𝑥∀∃!𝑦 ∈ 𝐴(𝜑 → 𝜓)
 
22-Jul-2026nfalseu 17294 Bound-variable hypothesis builder for "all some one". This is the "all some one" counterpart of nfals 17263. Unlike the set.mm version of this theorem, no disjoint variable condition is needed, because nfeu 2105 here does not require one. (Contributed by David A. Wheeler, 22-Jul-2026.)
Ⅎ𝑥𝜑    &   Ⅎ𝑥𝜓    ⇒   Ⅎ𝑥∀∃!𝑦(𝜑 → 𝜓)
 
22-Jul-2026ralseubii 17293 Congruence for "all some one" restricted to a class. This is the "all some one" counterpart of ralsbii 17261. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 ↔ 𝜒)    &   (𝜓 ↔ 𝜃)    ⇒   (∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃!𝑥 ∈ 𝐴(𝜒 → 𝜃))
 
22-Jul-2026alseubii 17292 Congruence: equivalents may be substituted inside an "all some one". This is the "all some one" counterpart of alsbii 17260. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 ↔ 𝜒)    &   (𝜓 ↔ 𝜃)    ⇒   (∀∃!𝑥(𝜑 → 𝜓) ↔ ∀∃!𝑥(𝜒 → 𝜃))
 
22-Jul-2026ralseu2d 17291 Deduction rule: Given "all some one" applied to a class, you can extract the "exactly one" part. Note that the witness must satisfy the antecedent 𝜓, not merely be a member of 𝐴. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 → ∀∃!𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   (𝜑 → ∃!𝑥 ∈ 𝐴 𝜓)
 
22-Jul-2026ralseu1d 17290 Deduction rule: Given "all some one" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 → ∀∃!𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))
 
22-Jul-2026alseu2d 17289 Deduction rule: Given "all some one" applied to a top-level inference, you can extract the "exactly one" part. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 → ∀∃!𝑥(𝜓 → 𝜒))    ⇒   (𝜑 → ∃!𝑥𝜓)
 
22-Jul-2026alseu1d 17288 Deduction rule: Given "all some one" applied to a top-level inference, you can extract the "for all" part. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 → ∀∃!𝑥(𝜓 → 𝜒))    ⇒   (𝜑 → ∀𝑥(𝜓 → 𝜒))
 
22-Jul-2026ralseud 17287 Introduction rule for "all some one" restricted to a class. This is the converse of ralseu1d 17290 and ralseu2d 17291 taken together. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))    &   (𝜑 → ∃!𝑥 ∈ 𝐴 𝜓)    ⇒   (𝜑 → ∀∃!𝑥 ∈ 𝐴(𝜓 → 𝜒))
 
22-Jul-2026alseud 17286 Introduction rule: "all some one" holds if the "for all" part holds and the antecedent has exactly one witness. This is the converse of alseu1d 17288 and alseu2d 17289 taken together. (Contributed by David A. Wheeler, 22-Jul-2026.)
(𝜑 → ∀𝑥(𝜓 → 𝜒))    &   (𝜑 → ∃!𝑥𝜓)    ⇒   (𝜑 → ∀∃!𝑥(𝜓 → 𝜒))
 
22-Jul-2026ralseurals 17285 "All some one" restricted to a class implies "all some" restricted to that class. Restricted counterpart of alseuals 17284. (Contributed by David A. Wheeler, 22-Jul-2026.)
(∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) → ∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓))
 
22-Jul-2026alseuals 17284 "All some one" implies "all some": requiring exactly one witness is stronger than requiring at least one. Any consequence of an allsome statement is therefore a consequence of the corresponding "all some one" statement, which is how alseu-no-surprise 17298 is proved. (Contributed by David A. Wheeler, 22-Jul-2026.)
(∀∃!𝑥(𝜑 → 𝜓) → ∀∃𝑥(𝜑 → 𝜓))
 
22-Jul-2026dfralseu2 17283 The bounded "all some one" form is the general form with the class membership folded into the antecedent. This is the "all some one" counterpart of dfrals2 17249. (Contributed by David A. Wheeler, 22-Jul-2026.)
(∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃!𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓))
 
22-Jul-2026df-ralseu 17282 Define "all some one" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and exactly one 𝑥 in 𝐴 satisfies 𝜑. (Contributed by David A. Wheeler, 22-Jul-2026.)
(∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∃!𝑥 ∈ 𝐴 𝜑))
 
22-Jul-2026df-alseu 17281 Define "all some one" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true and exactly one 𝑥 satisfies 𝜑. (Contributed by David A. Wheeler, 22-Jul-2026.)
(∀∃!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑))
 
22-Jul-2026wralseu 17280 Extend wff definition to include "all some one" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and exactly one 𝑥 in 𝐴 satisfies 𝜑. (Contributed by David A. Wheeler, 22-Jul-2026.)
wff ∀∃!𝑥 ∈ 𝐴(𝜑 → 𝜓)
 
22-Jul-2026walseu 17279 Extend wff definition to include "all some one" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true, and exactly one 𝑥 satisfies 𝜑. (Contributed by David A. Wheeler, 22-Jul-2026.)
wff ∀∃!𝑥(𝜑 → 𝜓)
 
22-Jul-2026mptmex 5945 If a function given by maps-to notation is inhabited, then the class it is defined on is inhabited. (Contributed by Jim Kingdon, 22-Jul-2026.)
(𝐶 ∈ (𝑥 ∈ 𝐴 ↦ 𝐵) → ∃𝑦 𝑦 ∈ 𝐴)
 
20-Jul-20262alsraln0idm 17278 Nested general "all some" quantifiers with class membership as their antecedents, for the same class 𝐴: 𝜑 holds for every 𝑥 and every 𝑦 in 𝐴, and 𝐴 is inhabited. (Contributed by Peter Mazsa, 28-May-2019.) (Revised by David A. Wheeler, 20-Jul-2026.)
(∀∃𝑥(𝑥 ∈ 𝐴 → ∀∃𝑦(𝑦 ∈ 𝐴 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴))
 
20-Jul-20262alsraln0m 17277 Nested general "all some" quantifiers with class membership as their antecedents: 𝜑 holds for every 𝑥 in 𝐴 and every 𝑦 in 𝐵, and both 𝐴 and 𝐵 are inhabited. (Contributed by Peter Mazsa, 28-May-2019.) (Revised by David A. Wheeler, 20-Jul-2026.)
(∀∃𝑥(𝑥 ∈ 𝐴 → ∀∃𝑦(𝑦 ∈ 𝐵 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ∧ (∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑦 𝑦 ∈ 𝐵)))
 
20-Jul-2026n0alsm 17276 If 𝐴 is inhabited, then the general "all some" quantifier with class membership as its antecedent reduces to the assertion that 𝜑 holds for every 𝑥 in 𝐴. (Contributed by Peter Mazsa, 19-Dec-2018.) (Revised by David A. Wheeler, 20-Jul-2026.)
(∃𝑥 𝑥 ∈ 𝐴 → (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥 ∈ 𝐴 𝜑))
 
20-Jul-2026alsraln0m 17273 The general "all some" quantifier with class membership as its antecedent holds if and only if 𝜑 holds for every 𝑥 in 𝐴 and 𝐴 is inhabited. This is the intuitionistic form of what set.mm states using 𝐴 ≠ ∅; see the section comment. (Contributed by Peter Mazsa, 28-Nov-2018.) (Revised by David A. Wheeler, 20-Jul-2026.)
(∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴))
 
20-Jul-2026alsralrex 17272 The general "all some" quantifier with class membership as its antecedent holds if and only if 𝜑 holds for every 𝑥 in 𝐴 and some 𝑥 in 𝐴 satisfies 𝜑. (Contributed by Peter Mazsa, 27-Nov-2018.) (Revised by David A. Wheeler, 20-Jul-2026.)
(∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 ∈ 𝐴 𝜑))
 
20-Jul-2026ralsanmo 17271 An "all some" statement restricted to a class, conjoined with the claim that at most one 𝑥 in 𝐴 satisfies its antecedent, is equivalent to the universal part conjoined with the claim that exactly one 𝑥 in 𝐴 satisfies the antecedent. This is the restricted counterpart of alsanmo 17270. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
((∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∃*𝑥 ∈ 𝐴 𝜑) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∃!𝑥 ∈ 𝐴 𝜑))
 
20-Jul-2026alsanmo 17270 An "all some" statement conjoined with the claim that at most one 𝑥 satisfies its antecedent is equivalent to the universal part conjoined with the claim that exactly one 𝑥 satisfies the antecedent. The "all some" quantifier supplies the existence of such an 𝑥 and ∃*𝑥𝜑 supplies the at-most-one part, so together they yield ∃!𝑥𝜑. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
((∀∃𝑥(𝜑 → 𝜓) ∧ ∃*𝑥𝜑) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃!𝑥𝜑))
 
20-Jul-2026rexrals 17269 If a member of 𝐴 satisfying the antecedent exists, then a restricted "all some" statement reduces to its universal part. This is the restricted counterpart of rexals 17275. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
(∃𝑥 ∈ 𝐴 𝜑 → (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀𝑥 ∈ 𝐴 (𝜑 → 𝜓)))
 
20-Jul-2026ralrals 17268 If the universal part of a restricted "all some" statement holds, then the statement reduces to the existence of a member of 𝐴 satisfying its antecedent. This is the restricted counterpart of ralals 17274. (Contributed by Peter Mazsa and David A. Wheeler, 20-Jul-2026.)
(∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) → (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∃𝑥 ∈ 𝐴 𝜑))
 
20-Jul-2026als-no-surprise 17266 Demonstrate that there is never a "surprise" when using the allsome quantifier, that is, it is never possible for the consequent to be both always true and always false. This uses the definition of df-als 17247: the universal parts give ∀𝑥¬ 𝜑, which contradicts the witness that the allsome quantifier supplies. Ordinary "for all" with implication has no such property, since ∀𝑥(𝜑 → 𝜓) and ∀𝑥(𝜑 → ¬ 𝜓) can both hold when nothing satisfies 𝜑. (Contributed by David A. Wheeler, 27-Oct-2018.) (Revised by David A. Wheeler, 20-Jul-2026.)
¬ (∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓))
 
20-Jul-2026ralsmd 17257 Deduction rule: Given "all some" applied to a class, the class is inhabited. This is stronger than ralsn0d 17256, which only concludes that the class is nonempty; see n0r 3535. (Contributed by David A. Wheeler, 20-Jul-2026.)
(𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   (𝜑 → ∃𝑥 𝑥 ∈ 𝐴)
 
19-Jul-2026disjdifg 3598 A class and its relative complement are disjoint. (Contributed by NM, 24-Mar-1998.) Generalize from disjdif 3599. (Revised by BJ, 19-Jul-2026.)
(𝐴 ⊆ 𝐵 → (𝐴 ∩ (𝐶 ∖ 𝐵)) = ∅)
 
19-Jul-2026sseq0b 3564 The only subclass of the empty class is itself. (Contributed by NM, 7-Mar-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) Strengthen sseq0 3565 to a biconditional. (Revised by BJ, 19-Jul-2026.)
(𝐴 = ∅ → (𝐵 ⊆ 𝐴 ↔ 𝐵 = ∅))
 
18-Jul-2026sepab 4278 Separation Scheme (Aussonderung) in terms of a class abstraction. Prefer using the more natural statement rabexg 4279. (Contributed by NM, 8-Jun-1994.) Put in closed form. (Revised by BJ, 18-Jul-2026.)
(𝐴 ∈ 𝑉 → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∈ V)
 
18-Jul-2026inssdif0im 3592 Intersection, subclass, and difference relationship. The converse holds in classical logic but not in intuitionistic logic. (Contributed by Jim Kingdon, 3-Aug-2018.) (Proof shortened by BJ, 18-Jul-2026.)
((𝐴 ∩ 𝐵) ⊆ 𝐶 → (𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅)
 
15-Jul-2026rexals 17275 If some 𝑥 in 𝐴 satisfies 𝜑, then the general "all some" quantifier with class membership as its antecedent reduces to the assertion that 𝜑 holds for every 𝑥 in 𝐴. See rexrals 17269 for the restricted counterpart. (Contributed by Peter Mazsa, 19-Dec-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
(∃𝑥 ∈ 𝐴 𝜑 → (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥 ∈ 𝐴 𝜑))
 
15-Jul-2026ralals 17274 If 𝜑 holds for every 𝑥 in 𝐴, then the general "all some" quantifier with class membership as its antecedent reduces to the assertion that some 𝑥 in 𝐴 satisfies 𝜑. See ralrals 17268 for the restricted counterpart. (Contributed by Peter Mazsa, 19-Dec-2018.) (Revised by David A. Wheeler, 15-Jul-2026.)
(∀𝑥 ∈ 𝐴 𝜑 → (∀∃𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∃𝑥 ∈ 𝐴 𝜑))
 
14-Jul-2026uniex2 4581 The Axiom of Union using the standard abbreviation for union. Given any set 𝑥, its union 𝑦 exists. (Contributed by NM, 4-Jun-2006.) (Proof shortened by BJ, 14-Jul-2026.)
∃𝑦 𝑦 = ∪ 𝑥
 
14-Jul-2026sepgi 4252 Inference associated with sepg 4251. (Contributed by NM, 21-Jun-1993.) (Revised by BJ, 14-Jul-2026.)
𝐴 ∈ V    ⇒   ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
 
13-Jul-2026f1setfi 7317 The set of injections between two finite sets is finite. (Contributed by Jim Kingdon, 13-Jul-2026.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → {𝑓 ∣ 𝑓:𝐴–1-1→𝐵} ∈ Fin)
 
13-Jul-2026fdcf1 7316 It is decidable whether a function from a finite set into another finite set is one-to-one. (Contributed by Jim Kingdon, 13-Jul-2026.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐹:𝐴⟶𝐵) → DECID 𝐹:𝐴–1-1→𝐵)
 
12-Jul-2026rals-no-surprise 17267 Demonstrate that there is never a "surprise" when using the allsome quantifier restricted to a class, that is, it is never possible for the consequent to be both always true and always false of the members of 𝐴 that satisfy the antecedent. This is the restricted counterpart of als-no-surprise 17266, and follows from it by dfrals2 17249. Note that this holds without any assumption that 𝐴 is inhabited; that is the point of allsome, since the corresponding claim for the ordinary restricted "for all" fails when nothing in 𝐴 satisfies 𝜑. (Contributed by David A. Wheeler, 12-Jul-2026.)
¬ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∀∃𝑥 ∈ 𝐴(𝜑 → ¬ 𝜓))
 
12-Jul-2026cbvals 17265 Rule used to change bound variables, using implicit substitution. (Contributed by David A. Wheeler, 12-Jul-2026.)
(𝑥 = 𝑦 → (𝜑 ↔ 𝜒))    &   (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))    ⇒   (∀∃𝑥(𝜑 → 𝜓) ↔ ∀∃𝑦(𝜒 → 𝜃))
 
12-Jul-2026nfrals 17264 Bound-variable hypothesis builder for "all some" restricted to a class. (Contributed by David A. Wheeler, 12-Jul-2026.)
Ⅎ𝑥𝐴    &   Ⅎ𝑥𝜑    &   Ⅎ𝑥𝜓    ⇒   Ⅎ𝑥∀∃𝑦 ∈ 𝐴(𝜑 → 𝜓)
 
12-Jul-2026nfals 17263 Bound-variable hypothesis builder for "all some". (Contributed by David A. Wheeler, 12-Jul-2026.)
Ⅎ𝑥𝜑    &   Ⅎ𝑥𝜓    ⇒   Ⅎ𝑥∀∃𝑦(𝜑 → 𝜓)
 
12-Jul-2026alsbid 17262 Deduction form of alsbii 17260. (Contributed by David A. Wheeler, 12-Jul-2026.)
Ⅎ𝑥𝜑    &   (𝜑 → (𝜓 ↔ 𝜃))    &   (𝜑 → (𝜒 ↔ 𝜏))    ⇒   (𝜑 → (∀∃𝑥(𝜓 → 𝜒) ↔ ∀∃𝑥(𝜃 → 𝜏)))
 
12-Jul-2026ralsbii 17261 Congruence for "all some" restricted to a class. (Contributed by David A. Wheeler, 12-Jul-2026.)
(𝜑 ↔ 𝜒)    &   (𝜓 ↔ 𝜃)    ⇒   (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃𝑥 ∈ 𝐴(𝜒 → 𝜃))
 
12-Jul-2026alsbii 17260 Congruence: equivalents may be substituted inside an "all some". (Contributed by David A. Wheeler, 12-Jul-2026.)
(𝜑 ↔ 𝜒)    &   (𝜓 ↔ 𝜃)    ⇒   (∀∃𝑥(𝜑 → 𝜓) ↔ ∀∃𝑥(𝜒 → 𝜃))
 
12-Jul-2026ralsex 17259 The consequent of an "all some" restricted to a class is witnessed: some member of 𝐴 satisfying 𝜑 also satisfies 𝜓. Restricted counterpart of alsex 17258. (Contributed by David A. Wheeler, 12-Jul-2026.)
(∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) → ∃𝑥 ∈ 𝐴 𝜓)
 
12-Jul-2026alsex 17258 The consequent of an "all some" is witnessed: if 𝜓 holds of every 𝑥 satisfying 𝜑, and some 𝑥 satisfies 𝜑, then some 𝑥 satisfies 𝜓. This is the positive counterpart of als-no-surprise 17266, and it is the property that ordinary "for all" with implication lacks: from ∀𝑥(𝜑 → 𝜓) alone nothing whatever follows about 𝜓, since nothing need satisfy 𝜑. It is the allsome quantifier says what a speaker of "all Martians are green" usually means. (Contributed by David A. Wheeler, 12-Jul-2026.)
(∀∃𝑥(𝜑 → 𝜓) → ∃𝑥𝜓)
 
12-Jul-2026ralsn0d 17256 Deduction rule: Given "all some" applied to a class, the class is not the empty set. (Contributed by David A. Wheeler, 23-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
(𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   (𝜑 → 𝐴 ≠ ∅)
 
12-Jul-2026rals2d 17255 Deduction rule: Given "all some" applied to a class, you can extract the "there exists" part. Note that the witness must satisfy the antecedent 𝜓, not merely be a member of 𝐴. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
(𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   (𝜑 → ∃𝑥 ∈ 𝐴 𝜓)
 
12-Jul-2026rals1d 17254 Deduction rule: Given "all some" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
(𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))    ⇒   (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))
 
12-Jul-2026ralsd 17251 Introduction rule for "all some" restricted to a class. This is the converse of rals1d 17254 and rals2d 17255 taken together. (Contributed by David A. Wheeler, 12-Jul-2026.)
(𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))    &   (𝜑 → ∃𝑥 ∈ 𝐴 𝜓)    ⇒   (𝜑 → ∀∃𝑥 ∈ 𝐴(𝜓 → 𝜒))
 
12-Jul-2026alsd 17250 Introduction rule: "all some" holds if the "for all" part holds and the antecedent has a witness. This is the converse of als1d 17252 and als2d 17253 taken together, and is what lets an "all some" statement be proved rather than merely taken apart. (Contributed by David A. Wheeler, 12-Jul-2026.)
(𝜑 → ∀𝑥(𝜓 → 𝜒))    &   (𝜑 → ∃𝑥𝜓)    ⇒   (𝜑 → ∀∃𝑥(𝜓 → 𝜒))
 
12-Jul-2026dfrals2 17249 The bounded "all some" form is the general form with the class membership folded into the antecedent. (Contributed by David A. Wheeler, 22-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
(∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓))
 
12-Jul-2026df-rals 17248 Define "all some" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and there is at least one 𝑥 in 𝐴 where 𝜑 is true.

An older definition of the "all some" quantifier when scoped to a class, named df-alsc and now removed, instead applied a bare formula 𝜑 to the members of a class, asserting only (∀𝑥 ∈ 𝐴𝜑 ∧ ∃𝑥𝑥 ∈ 𝐴), that is, that the formula held throughout 𝐴 and that 𝐴 had at least one member. I've now decided that that was a mistake. Its older existence conjunct ∃𝑥𝑥 ∈ 𝐴 did not require any member of 𝐴 to satisfy the antecedent, so if the formula was itself an implication, that inner implication could still be vacuously true, which is precisely what the allsome quantifier exists to prevent. For example, the older definition meant that "among Martians, all tall ones are green" could be considered true if there are Martians, but no tall Martians. This version of the definition instead ensures that claims of the form "among Martians, all tall ones are green" can only be true if all tall Martians are green and that there is at least one tall Martian. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)

(∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∃𝑥 ∈ 𝐴 𝜑))
 
12-Jul-2026wrals 17246 Extend wff definition to include "all some" applied to a class, which means 𝜓 is true whenever 𝜑 is true for 𝑥 in 𝐴, and there is at least one 𝑥 in 𝐴 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
wff ∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓)
 
12-Jul-2026wals 17245 Extend wff definition to include "all some" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true, and there is at least one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.) (Revised by David A. Wheeler, 12-Jul-2026.)
wff ∀∃𝑥(𝜑 → 𝜓)
 
12-Jul-2026vvin 3569 Two classes are both the universal class if and only if their intersection is the universal class. Dual of un00 3567. (Contributed by BJ, 12-Jul-2026.)
((𝐴 = V ∧ 𝐵 = V) ↔ (𝐴 ∩ 𝐵) = V)
 
7-Jul-2026cmnsubm 14161 A submonoid of a commutative monoid is commutative. (Contributed by Jim Kingdon, 7-Jul-2026.)
(𝜑 → 𝑆 ∈ (SubMnd‘𝐺))    &   (𝜑 → 𝐺 ∈ CMnd)    &   𝐻 = (𝐺 ↾s 𝑆)    ⇒   (𝜑 → 𝐻 ∈ CMnd)
 
29-Jun-2026dichmul0or 16879 Real number dichotomy is equivalent to the zero product principle for complex numbers: if a product is zero, one of its factors must be zero. (Contributed by Matthew House, 29-Jun-2026.)
(∀𝑥 ∈ ℝ ∀𝑦 ∈ ℝ (𝑥 ≤ 𝑦 ∨ 𝑦 ≤ 𝑥) ↔ ∀𝑧 ∈ ℂ ∀𝑤 ∈ ℂ ((𝑧 · 𝑤) = 0 → (𝑧 = 0 ∨ 𝑤 = 0)))
 
29-Jun-2026dichmul0orlem5 16876 Lemma for dichmul0or 16879. (Contributed by Matthew House, 29-Jun-2026.)
(𝜑 → 𝐴 ∈ ℝ)    &   (𝜑 → ((abs‘𝐴) + 𝐴) = 0)    ⇒   (𝜑 → 𝐴 ≤ 0)
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