P. 470 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 46901-47000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	tworepnotupword 46901	Concatenation of identical singletons is never an increasing sequence. (Contributed by Ender Ting, 22-Nov-2024.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ¬ (⟨“𝐴”⟩ ++ ⟨“𝐴”⟩) ∈ UpWord 𝑆
Theorem	upwrdfi 46902*	There is a finite number of strictly increasing sequences of a given length over finite alphabet. Trivially holds for invalid lengths where there're zero matching sequences. (Contributed by Ender Ting, 5-Jan-2024.)
⊢ (𝑆 ∈ Fin → {𝑎 ∈ UpWord 𝑆 ∣ (♯‘𝑎) = 𝑇} ∈ Fin)
21.45.2  Scratchpad for number theory
Theorem	evenwodadd 46903	If an integer is multiplied by its sum with an odd number (thus changing its parity), the result is even. (Contributed by Ender Ting, 30-Apr-2025.)
⊢ (𝜑 → 𝑖 ∈ ℤ)    &   ⊢ (𝜑 → 𝑗 ∈ ℤ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑗)    ⇒   ⊢ (𝜑 → 2 ∥ (𝑖 · (𝑖 + 𝑗)))
21.46  Mathbox for Jarvin Udandy
Theorem	hirstL-ax3 46904	The third axiom of a system called "L" but proven to be a theorem since set.mm uses a different third axiom. This is named hirst after Holly P. Hirst and Jeffry L. Hirst. Axiom A3 of [Mendelson] p. 35. (Contributed by Jarvin Udandy, 7-Feb-2015.) (Proof modification is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → ((¬ 𝜑 → 𝜓) → 𝜑))
Theorem	ax3h 46905	Recover ax-3 8 from hirstL-ax3 46904. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑))
Theorem	aibandbiaiffaiffb 46906	A closed form showing (a implies b and b implies a) same-as (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ (𝜑 ↔ 𝜓))
Theorem	aibandbiaiaiffb 46907	A closed form showing (a implies b and b implies a) implies (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) → (𝜑 ↔ 𝜓))
Theorem	notatnand 46908	Do not use. Use intnanr instead. Given not a, there exists a proof for not (a and b). (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    ⇒   ⊢ ¬ (𝜑 ∧ 𝜓)
Theorem	aistia 46909	Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    ⇒   ⊢ 𝜑
Theorem	aisfina 46910	Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    ⇒   ⊢ ¬ 𝜑
Theorem	bothtbothsame 46911	Given both a, b are equivalent to ⊤, there exists a proof for a is the same as b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ↔ 𝜓)
Theorem	bothfbothsame 46912	Given both a, b are equivalent to ⊥, there exists a proof for a is the same as b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ (𝜑 ↔ 𝜓)
Theorem	aiffbbtat 46913	Given a is equivalent to b, b is equivalent to ⊤ there exists a proof for a is equivalent to T. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ↔ ⊤)
Theorem	aisbbisfaisf 46914	Given a is equivalent to b, b is equivalent to ⊥ there exists a proof for a is equivalent to F. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	axorbtnotaiffb 46915	Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1512 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    ⇒   ⊢ ¬ (𝜑 ↔ 𝜓)
Theorem	aiffnbandciffatnotciffb 46916	Given a is equivalent to (not b), c is equivalent to a, there exists a proof for ( not ( c iff b ) ). (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ ¬ 𝜓)    &   ⊢ (𝜒 ↔ 𝜑)    ⇒   ⊢ ¬ (𝜒 ↔ 𝜓)
Theorem	axorbciffatcxorb 46917	Given a is equivalent to (not b), c is equivalent to a. there exists a proof for ( c xor b ). (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    &   ⊢ (𝜒 ↔ 𝜑)    ⇒   ⊢ (𝜒 ⊻ 𝜓)
Theorem	aibnbna 46918	Given a implies b, (not b), there exists a proof for (not a). (Contributed by Jarvin Udandy, 1-Sep-2016.)
⊢ (𝜑 → 𝜓)    &   ⊢ ¬ 𝜓    ⇒   ⊢ ¬ 𝜑
Theorem	aibnbaif 46919	Given a implies b, not b, there exists a proof for a is F. (Contributed by Jarvin Udandy, 1-Sep-2016.)
⊢ (𝜑 → 𝜓)    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	aiffbtbat 46920	Given a is equivalent to b, T. is equivalent to b. there exists a proof for a is equivalent to T. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ (⊤ ↔ 𝜓)    ⇒   ⊢ (𝜑 ↔ ⊤)
Theorem	astbstanbst 46921	Given a is equivalent to T., also given that b is equivalent to T, there exists a proof for a and b is equivalent to T. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ ((𝜑 ∧ 𝜓) ↔ ⊤)
Theorem	aistbistaandb 46922	Given a is equivalent to T., also given that b is equivalent to T, there exists a proof for (a and b). (Contributed by Jarvin Udandy, 9-Sep-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ∧ 𝜓)
Theorem	aisbnaxb 46923	Given a is equivalent to b, there exists a proof for (not (a xor b)). (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ (𝜑 ↔ 𝜓)    ⇒   ⊢ ¬ (𝜑 ⊻ 𝜓)
Theorem	atbiffatnnb 46924	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 46925	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 46926	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 46927	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 46928	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 46929	Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    &   ⊢ (𝜑 → 𝜓)    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	conimpfalt 46930	Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    &   ⊢ (𝜑 → 𝜓)    ⇒   ⊢ (𝜑 ↔ ⊥)
Theorem	aistbisfiaxb 46931	Given a is equivalent to T., Given b is equivalent to F. there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	aisfbistiaxb 46932	Given a is equivalent to F., Given b is equivalent to T., there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	aifftbifffaibif 46933	Given a is equivalent to T., Given b is equivalent to F., there exists a proof for that a implies b is false. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ ((𝜑 → 𝜓) ↔ ⊥)
Theorem	aifftbifffaibifff 46934	Given a is equivalent to T., Given b is equivalent to F., there exists a proof for that a iff b is false. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (𝜑 ↔ ⊤)    &   ⊢ (𝜓 ↔ ⊥)    ⇒   ⊢ ((𝜑 ↔ 𝜓) ↔ ⊥)
Theorem	atnaiana 46935	Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ 𝜑    ⇒   ⊢ ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑))
Theorem	ainaiaandna 46936	Given a, a implies it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ 𝜑    ⇒   ⊢ (𝜑 → ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑)))
Theorem	abcdta 46937	Given (((a and b) and c) and d), there exists a proof for a. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜑
Theorem	abcdtb 46938	Given (((a and b) and c) and d), there exists a proof for b. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜓
Theorem	abcdtc 46939	Given (((a and b) and c) and d), there exists a proof for c. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜒
Theorem	abcdtd 46940	Given (((a and b) and c) and d), there exists a proof for d. (Contributed by Jarvin Udandy, 3-Sep-2016.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃)    ⇒   ⊢ 𝜃
Theorem	abciffcbatnabciffncba 46941	Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. Closed form. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) → ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑))
Theorem	abciffcbatnabciffncbai 46942	Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜒 ∧ 𝜓) ∧ 𝜑))    ⇒   ⊢ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) → ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑))
Theorem	nabctnabc 46943	not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ ¬ (𝜑 → (𝜓 ∧ 𝜒))    ⇒   ⊢ (¬ 𝜑 → (𝜓 ∧ 𝜒))
Theorem	jabtaib 46944	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 46945	From one negated implication it is not the case its nonnegated form and a random others are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
⊢ ¬ (𝜑 → 𝜓)    ⇒   ⊢ ¬ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜃))
Theorem	twonotinotbothi 46946	From these two negated implications it is not the case their nonnegated forms are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
⊢ ¬ (𝜑 → 𝜓)    &   ⊢ ¬ (𝜒 → 𝜃)    ⇒   ⊢ ¬ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜃))
Theorem	clifte 46947	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ (𝜑 ∧ ¬ 𝜒)    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓 ∧ 𝜒)))
Theorem	cliftet 46948	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ (𝜑 ∧ 𝜒)    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))
Theorem	clifteta 46949	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓 ∧ 𝜒))    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓 ∧ 𝜒)))
Theorem	cliftetb 46950	show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
⊢ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ ¬ 𝜒))    &   ⊢ 𝜃    ⇒   ⊢ (𝜃 ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))
Theorem	confun 46951	Given the hypotheses there exists a proof for (c implies ( d iff a ) ). (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ 𝜑    &   ⊢ (𝜒 → 𝜓)    &   ⊢ (𝜒 → 𝜃)    &   ⊢ (𝜑 → (𝜑 → 𝜓))    ⇒   ⊢ (𝜒 → (𝜃 ↔ 𝜑))
Theorem	confun2 46952	Confun simplified to two propositions. (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ (𝜓 → 𝜑)    &   ⊢ (𝜓 → ¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)))    &   ⊢ ((𝜓 → 𝜑) → ((𝜓 → 𝜑) → 𝜑))    ⇒   ⊢ (𝜓 → (¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)) ↔ (𝜓 → 𝜑)))
Theorem	confun3 46953	Confun's more complex form where both a,d have been "defined". (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ (𝜑 ↔ (𝜒 → 𝜓))    &   ⊢ (𝜃 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ (𝜒 → 𝜓)    &   ⊢ (𝜒 → ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ ((𝜒 → 𝜓) → ((𝜒 → 𝜓) → 𝜓))    ⇒   ⊢ (𝜒 → (¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)) ↔ (𝜒 → 𝜓)))
Theorem	confun4 46954	An attempt at derivative. Resisted simplest path to a proof. (Contributed by Jarvin Udandy, 6-Sep-2020.)
⊢ 𝜑    &   ⊢ ((𝜑 → 𝜓) → 𝜓)    &   ⊢ (𝜓 → (𝜑 → 𝜒))    &   ⊢ ((𝜒 → 𝜃) → ((𝜑 → 𝜃) ↔ 𝜓))    &   ⊢ (𝜏 ↔ (𝜒 → 𝜃))    &   ⊢ (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ 𝜓    &   ⊢ (𝜒 → 𝜃)    ⇒   ⊢ (𝜒 → (𝜓 → 𝜏))
Theorem	confun5 46955	An attempt at derivative. Resisted simplest path to a proof. Interesting that ch, th, ta, et were all provable. (Contributed by Jarvin Udandy, 7-Sep-2020.)
⊢ 𝜑    &   ⊢ ((𝜑 → 𝜓) → 𝜓)    &   ⊢ (𝜓 → (𝜑 → 𝜒))    &   ⊢ ((𝜒 → 𝜃) → ((𝜑 → 𝜃) ↔ 𝜓))    &   ⊢ (𝜏 ↔ (𝜒 → 𝜃))    &   ⊢ (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ⊢ 𝜓    &   ⊢ (𝜒 → 𝜃)    ⇒   ⊢ (𝜒 → (𝜂 ↔ 𝜏))
Theorem	plcofph 46956	Given, a,b and a "definition" for c, c is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜒 ↔ ((((𝜑 ∧ 𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   ⊢ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ 𝜒
Theorem	pldofph 46957	Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜏 ↔ ((𝜒 → 𝜃) ∧ (𝜑 ↔ 𝜒) ∧ ((𝜑 → 𝜓) → (𝜓 ↔ 𝜃))))    &   ⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜒    &   ⊢ 𝜃    ⇒   ⊢ 𝜏
Theorem	plvcofph 46958	Given, a,b,d, and "definitions" for c, e, f: f is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜒 ↔ ((((𝜑 ∧ 𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   ⊢ (𝜏 ↔ ((𝜒 → 𝜃) ∧ (𝜑 ↔ 𝜒) ∧ ((𝜑 → 𝜓) → (𝜓 ↔ 𝜃))))    &   ⊢ (𝜂 ↔ (𝜒 ∧ 𝜏))    &   ⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜃    ⇒   ⊢ 𝜂
Theorem	plvcofphax 46959	Given, a,b,d, and "definitions" for c, e, f, g: g is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
⊢ (𝜒 ↔ ((((𝜑 ∧ 𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   ⊢ (𝜏 ↔ ((𝜒 → 𝜃) ∧ (𝜑 ↔ 𝜒) ∧ ((𝜑 → 𝜓) → (𝜓 ↔ 𝜃))))    &   ⊢ (𝜂 ↔ (𝜒 ∧ 𝜏))    &   ⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜃    &   ⊢ (𝜁 ↔ ¬ (𝜓 ∧ ¬ 𝜏))    ⇒   ⊢ 𝜁
Theorem	plvofpos 46960	rh is derivable because ONLY one of ch, th, ta, et is implied by mu. (Contributed by Jarvin Udandy, 11-Sep-2020.)
⊢ (𝜒 ↔ (¬ 𝜑 ∧ ¬ 𝜓))    &   ⊢ (𝜃 ↔ (¬ 𝜑 ∧ 𝜓))    &   ⊢ (𝜏 ↔ (𝜑 ∧ ¬ 𝜓))    &   ⊢ (𝜂 ↔ (𝜑 ∧ 𝜓))    &   ⊢ (𝜁 ↔ (((((¬ ((𝜇 → 𝜒) ∧ (𝜇 → 𝜃)) ∧ ¬ ((𝜇 → 𝜒) ∧ (𝜇 → 𝜏))) ∧ ¬ ((𝜇 → 𝜒) ∧ (𝜒 → 𝜂))) ∧ ¬ ((𝜇 → 𝜃) ∧ (𝜇 → 𝜏))) ∧ ¬ ((𝜇 → 𝜃) ∧ (𝜇 → 𝜂))) ∧ ¬ ((𝜇 → 𝜏) ∧ (𝜇 → 𝜂))))    &   ⊢ (𝜎 ↔ (((𝜇 → 𝜒) ∨ (𝜇 → 𝜃)) ∨ ((𝜇 → 𝜏) ∨ (𝜇 → 𝜂))))    &   ⊢ (𝜌 ↔ (𝜁 ∧ 𝜎))    &   ⊢ 𝜁    &   ⊢ 𝜎    ⇒   ⊢ 𝜌
Theorem	mdandyv0 46961	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv1 46962	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv2 46963	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv3 46964	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv4 46965	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv5 46966	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv6 46967	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv7 46968	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊥)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜑))
Theorem	mdandyv8 46969	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv9 46970	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv10 46971	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv11 46972	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜑)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv12 46973	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv13 46974	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜑)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv14 46975	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊥)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜑) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyv15 46976	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ ⊥)    &   ⊢ (𝜓 ↔ ⊤)    &   ⊢ (𝜒 ↔ ⊤)    &   ⊢ (𝜃 ↔ ⊤)    &   ⊢ (𝜏 ↔ ⊤)    &   ⊢ (𝜂 ↔ ⊤)    ⇒   ⊢ ((((𝜒 ↔ 𝜓) ∧ (𝜃 ↔ 𝜓)) ∧ (𝜏 ↔ 𝜓)) ∧ (𝜂 ↔ 𝜓))
Theorem	mdandyvr0 46977	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr1 46978	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr2 46979	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr3 46980	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr4 46981	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr5 46982	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr6 46983	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr7 46984	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜁))
Theorem	mdandyvr8 46985	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr9 46986	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr10 46987	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr11 46988	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr12 46989	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr13 46990	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr14 46991	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜁) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvr15 46992	Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ↔ 𝜁)    &   ⊢ (𝜓 ↔ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ↔ 𝜎) ∧ (𝜃 ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (𝜂 ↔ 𝜎))
Theorem	mdandyvrx0 46993	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx1 46994	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx2 46995	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx3 46996	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜑)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx4 46997	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx5 46998	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜑)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx6 46999	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜑)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜁))
Theorem	mdandyvrx7 47000	Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜁)    &   ⊢ (𝜓 ⊻ 𝜎)    &   ⊢ (𝜒 ↔ 𝜓)    &   ⊢ (𝜃 ↔ 𝜓)    &   ⊢ (𝜏 ↔ 𝜓)    &   ⊢ (𝜂 ↔ 𝜑)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜁))