P. 480 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 47901-48000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	tmachlem-agreesn 47901*	Scans for all tapes of a single agreement set are identical. (Contributed by Ender Ting, 28-Jul-2026.)
⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝐼 ∈ V)    &   ⊢ (𝜑 → 𝑇 = (𝑈 ↑m 𝐼))    &   ⊢ (𝜑 → 𝑆:𝑇⟶(𝒫 𝐼 ∩ Fin))    &   ⊢ (𝜑 → 𝐴 = (𝑧 ∈ 𝑇 ↦ {𝑦 ∈ 𝑇 ∣ (𝑦 ↾ (𝑆‘𝑧)) = (𝑧 ↾ (𝑆‘𝑧))}))    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑇 ∀𝑦 ∈ (𝐴‘𝑧)(𝑆‘𝑦) = (𝑆‘𝑧))    ⇒   ⊢ ((𝜑 ∧ 𝑎 ∈ 𝑇) → (𝑆 “ (𝐴‘𝑎)) = {(𝑆‘𝑎)})
Theorem	tmachlem-agreefin 47902*	Any agreement set has a finite (singleton) list of possible scans. (Contributed by Ender Ting, 28-Jul-2026.)
⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝐼 ∈ V)    &   ⊢ (𝜑 → 𝑇 = (𝑈 ↑m 𝐼))    &   ⊢ (𝜑 → 𝑆:𝑇⟶(𝒫 𝐼 ∩ Fin))    &   ⊢ (𝜑 → 𝐴 = (𝑧 ∈ 𝑇 ↦ {𝑦 ∈ 𝑇 ∣ (𝑦 ↾ (𝑆‘𝑧)) = (𝑧 ↾ (𝑆‘𝑧))}))    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑇 ∀𝑦 ∈ (𝐴‘𝑧)(𝑆‘𝑦) = (𝑆‘𝑧))    ⇒   ⊢ ((𝜑 ∧ 𝑏 ∈ ran 𝐴) → (𝑆 “ 𝑏) ∈ Fin)
Theorem	tmachlem-franscan 47903*	There is a finite number of different scan sets. (Contributed by Ender Ting, 28-Jul-2026.)
⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝐼 ∈ V)    &   ⊢ (𝜑 → 𝑇 = (𝑈 ↑m 𝐼))    &   ⊢ (𝜑 → 𝑆:𝑇⟶(𝒫 𝐼 ∩ Fin))    &   ⊢ (𝜑 → 𝐴 = (𝑧 ∈ 𝑇 ↦ {𝑦 ∈ 𝑇 ∣ (𝑦 ↾ (𝑆‘𝑧)) = (𝑧 ↾ (𝑆‘𝑧))}))    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑇 ∀𝑦 ∈ (𝐴‘𝑧)(𝑆‘𝑦) = (𝑆‘𝑧))    ⇒   ⊢ (𝜑 → ran 𝑆 ∈ Fin)
Theorem	tmachlem-fssscan 47904*	Any scan set is finite. (Contributed by Ender Ting, 28-Jul-2026.)
⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝐼 ∈ V)    &   ⊢ (𝜑 → 𝑇 = (𝑈 ↑m 𝐼))    &   ⊢ (𝜑 → 𝑆:𝑇⟶(𝒫 𝐼 ∩ Fin))    &   ⊢ (𝜑 → 𝐴 = (𝑧 ∈ 𝑇 ↦ {𝑦 ∈ 𝑇 ∣ (𝑦 ↾ (𝑆‘𝑧)) = (𝑧 ↾ (𝑆‘𝑧))}))    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑇 ∀𝑦 ∈ (𝐴‘𝑧)(𝑆‘𝑦) = (𝑆‘𝑧))    ⇒   ⊢ (𝜑 → ran 𝑆 ⊆ Fin)
Theorem	tmachfullfin 47905*	Folk theorem. For any algorithm deterministically processing a stream of data (essentially, an infinite tape with cell indices 𝐼 and finite alphabet 𝑈), if it terminates on every possible input, then it never looks beyond a finite portion ∪ ran 𝑆 of the input. Termination is expressed here with a weaker condition: that an execution may only look at a finite number of cells. Obviously, a program which finishes in a finite number of steps can only scan finite set of cells. This theorem has many corollaries, such as: any encoding scheme able to represent all integers has at least one non-decodable tape (in other terms, encoding of the infinity). I no longer have the source for this theorem but I believe I first read about it on LessWrong. My gratitude to Grok for suggesting that this theorem will require Axiom of Choice, and to DeepSeek for suggesting the topology-based proof route. (Contributed by Ender Ting, 28-Jul-2026.)
⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝐼 ∈ V)    &   ⊢ (𝜑 → 𝑇 = (𝑈 ↑m 𝐼))    &   ⊢ (𝜑 → 𝑆:𝑇⟶(𝒫 𝐼 ∩ Fin))    &   ⊢ (𝜑 → 𝐴 = (𝑧 ∈ 𝑇 ↦ {𝑦 ∈ 𝑇 ∣ (𝑦 ↾ (𝑆‘𝑧)) = (𝑧 ↾ (𝑆‘𝑧))}))    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑇 ∀𝑦 ∈ (𝐴‘𝑧)(𝑆‘𝑦) = (𝑆‘𝑧))    ⇒   ⊢ (𝜑 → ∪ ran 𝑆 ∈ Fin)
21.48  Mathbox for Jarvin Udandy
Theorem	hirstL-ax3 47906	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 47907	Recover ax-3 8 from hirstL-ax3 47906. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑))
Theorem	aibandbiaiffaiffb 47908	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 47909	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 47910	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 47911	Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊤)    ⇒   ⊢ 𝜑
Theorem	aisfina 47912	Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
⊢ (𝜑 ↔ ⊥)    ⇒   ⊢ ¬ 𝜑
Theorem	bothtbothsame 47913	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 47914	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 47915	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 47916	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 47917	Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1542 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
⊢ (𝜑 ⊻ 𝜓)    ⇒   ⊢ ¬ (𝜑 ↔ 𝜓)
Theorem	aiffnbandciffatnotciffb 47918	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 47919	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 47920	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 47921	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 47922	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 47923	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 47924	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 47925	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 47926	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 47927	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 47928	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 47929	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 47930	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 47931	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 47932	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 47933	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 47934	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 47935	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 47936	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 47937	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 47938	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 47939	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 47940	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 47941	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 47942	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 47943	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 47944	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 47945	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 47946	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 47947	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 47948	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 47949	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 47950	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 47951	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 47952	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 47953	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 47954	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 47955	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 47956	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 47957	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 47958	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 47959	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 47960	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 47961	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 47962	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 47963	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 47964	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 47965	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 47966	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 47967	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 47968	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 47969	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 47970	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 47971	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 47972	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 47973	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 47974	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 47975	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 47976	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 47977	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 47978	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 47979	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 47980	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 47981	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 47982	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 47983	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 47984	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 47985	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 47986	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 47987	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 47988	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 47989	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 47990	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 47991	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 47992	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 47993	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 47994	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 47995	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 47996	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 47997	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 47998	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 47999	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 48000	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))