P. 460 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 45901-46000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ax3h 45901	Recover ax-3 8 from hirstL-ax3 45900. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑))
Theorem	aibandbiaiffaiffb 45902	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 45903	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 45904	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 45905	Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    ⇒   ⊢ 𝜑
Theorem	aisfina 45906	Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    ⇒   ⊢ ¬ 𝜑
Theorem	bothtbothsame 45907	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 45908	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 45909	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 45910	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 45911	Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1508 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    ⇒   ⊢ ¬ (𝜑 ↔ 𝜓)
Theorem	aiffnbandciffatnotciffb 45912	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 45913	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 45914	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 45915	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 45916	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 45917	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 45918	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 45919	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 45920	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 45921	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 45922	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 45923	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 45924	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 45925	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 45926	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 45927	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 45928	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 45929	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 45930	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 45931	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 45932	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 45933	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 45934	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 45935	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 45936	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 45937	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 45938	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 45939	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 45940	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 45941	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 45942	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 45943	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 45944	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 45945	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 45946	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 45947	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 45948	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 45949	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 45950	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 45951	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 45952	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 45953	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 45954	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 45955	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 45956	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 45957	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 45958	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 45959	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 45960	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 45961	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 45962	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 45963	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 45964	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 45965	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 45966	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 45967	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 45968	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 45969	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 45970	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 45971	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 45972	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 45973	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 45974	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 45975	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 45976	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 45977	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 45978	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 45979	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 45980	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 45981	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 45982	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 45983	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 45984	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 45985	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 45986	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 45987	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 45988	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 45989	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 45990	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 45991	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 45992	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 45993	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 45994	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 45995	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 45996	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))
Theorem	mdandyvrx8 45997	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜎))
Theorem	mdandyvrx9 45998	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜎))
Theorem	mdandyvrx10 45999	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜎))
Theorem	mdandyvrx11 46000	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜁)) ∧ (𝜂 ⊻ 𝜎))