P. 421 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 42001-42100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	aibnbaif 42001	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 42002	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 42003	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 42004	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 42005	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 42006	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	bisaiaisb 42007	Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
Theorem	atbiffatnnbalt 42008	If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))
Theorem	abnotbtaxb 42009	Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ 𝜑    &   ⊢ ¬ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	abnotataxb 42010	Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
⊢ ¬ 𝜑    &   ⊢ 𝜓    ⇒   ⊢ (𝜑 ⊻ 𝜓)
Theorem	conimpf 42011	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 42012	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 42013	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 42014	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 42015	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 42016	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 42017	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 42018	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 42019	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 42020	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 42021	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 42022	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 42023	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 42024	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 42025	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 42026	For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
⊢ (𝜑 ∧ 𝜓)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	onenotinotbothi 42027	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 42028	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 42029	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 42030	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 42031	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 42032	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 42033	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 42034	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 42035	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 42036	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 42037	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 42038	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 42039	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 42040	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 42041	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 42042	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 42043	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 42044	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 42045	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 42046	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 42047	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 42048	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 42049	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 42050	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 42051	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 42052	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 42053	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 42054	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 42055	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 42056	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 42057	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 42058	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 42059	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 42060	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 42061	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 42062	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 42063	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 42064	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 42065	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 42066	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 42067	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 42068	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 42069	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 42070	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 42071	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 42072	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 42073	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 42074	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 42075	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 42076	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 42077	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 42078	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 42079	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 42080	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 42081	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 42082	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 42083	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 42084	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 42085	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 42086	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))
Theorem	mdandyvrx12 42087	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜎))
Theorem	mdandyvrx13 42088	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜎))
Theorem	mdandyvrx14 42089	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜁) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜎))
Theorem	mdandyvrx15 42090	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 ↔ 𝜓)    ⇒   ⊢ ((((𝜒 ⊻ 𝜎) ∧ (𝜃 ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜎)) ∧ (𝜂 ⊻ 𝜎))
Theorem	H15NH16TH15IH16 42091	Given 15 hypotheses and a 16th hypothesis, there exists a proof the 15 imply the 16th. (Contributed by Jarvin Udandy, 8-Sep-2016.)
⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜒    &   ⊢ 𝜃    &   ⊢ 𝜏    &   ⊢ 𝜂    &   ⊢ 𝜁    &   ⊢ 𝜎    &   ⊢ 𝜌    &   ⊢ 𝜇    &   ⊢ 𝜆    &   ⊢ 𝜅    &   ⊢ jph    &   ⊢ jps    &   ⊢ jch    &   ⊢ jth    ⇒   ⊢ (((((((((((((((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃) ∧ 𝜏) ∧ 𝜂) ∧ 𝜁) ∧ 𝜎) ∧ 𝜌) ∧ 𝜇) ∧ 𝜆) ∧ 𝜅) ∧ jph) ∧ jps) ∧ jch) → jth)
Theorem	dandysum2p2e4 42092	CONTRADICTION PROVED AT 1 + 1 = 2 . Given the right hypotheses we can prove a dandysum of 2+2=4. The qed step is the value '4' in Decimal BEING IMPLIED by the hypotheses. Note: Values that when added which exceed a 4bit value are not supported. Note: Digits begin from left (least) to right (greatest). e.g. 1000 would be '1', 0100 would be '2'. 0010 would be '4'. How to perceive the hypotheses' bits in order: ( th <-> F. ), ( ta <-> F. ) Would be input value X's first bit, and input value Y's first bit. ( et <-> F ), ( ze <-> F. ) would be input value X's second bit, and input value Y's second bit. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (𝜑 ↔ (𝜃 ∧ 𝜏))    &   ⊢ (𝜓 ↔ (𝜂 ∧ 𝜁))    &   ⊢ (𝜒 ↔ (𝜎 ∧ 𝜌))    &   ⊢ (𝜃 ↔ ⊥)    &   ⊢ (𝜏 ↔ ⊥)    &   ⊢ (𝜂 ↔ ⊤)    &   ⊢ (𝜁 ↔ ⊤)    &   ⊢ (𝜎 ↔ ⊥)    &   ⊢ (𝜌 ↔ ⊥)    &   ⊢ (𝜇 ↔ ⊥)    &   ⊢ (𝜆 ↔ ⊥)    &   ⊢ (𝜅 ↔ ((𝜃 ⊻ 𝜏) ⊻ (𝜃 ∧ 𝜏)))    &   ⊢ (jph ↔ ((𝜂 ⊻ 𝜁) ∨ 𝜑))    &   ⊢ (jps ↔ ((𝜎 ⊻ 𝜌) ∨ 𝜓))    &   ⊢ (jch ↔ ((𝜇 ⊻ 𝜆) ∨ 𝜒))    ⇒   ⊢ ((((((((((((((((𝜑 ↔ (𝜃 ∧ 𝜏)) ∧ (𝜓 ↔ (𝜂 ∧ 𝜁))) ∧ (𝜒 ↔ (𝜎 ∧ 𝜌))) ∧ (𝜃 ↔ ⊥)) ∧ (𝜏 ↔ ⊥)) ∧ (𝜂 ↔ ⊤)) ∧ (𝜁 ↔ ⊤)) ∧ (𝜎 ↔ ⊥)) ∧ (𝜌 ↔ ⊥)) ∧ (𝜇 ↔ ⊥)) ∧ (𝜆 ↔ ⊥)) ∧ (𝜅 ↔ ((𝜃 ⊻ 𝜏) ⊻ (𝜃 ∧ 𝜏)))) ∧ (jph ↔ ((𝜂 ⊻ 𝜁) ∨ 𝜑))) ∧ (jps ↔ ((𝜎 ⊻ 𝜌) ∨ 𝜓))) ∧ (jch ↔ ((𝜇 ⊻ 𝜆) ∨ 𝜒))) → ((((𝜅 ↔ ⊥) ∧ (jph ↔ ⊥)) ∧ (jps ↔ ⊤)) ∧ (jch ↔ ⊥)))
Theorem	mdandysum2p2e4 42093	CONTRADICTION PROVED AT 1 + 1 = 2 . Luckily Mario Carneiro did a successful version of his own. See Mario's Relevant Work: 1.3.14 Half adder and full adder in propositional calculus. Given the right hypotheses we can prove a dandysum of 2+2=4. The qed step is the value '4' in Decimal BEING IMPLIED by the hypotheses. Note: Values that when added which exceed a 4bit value are not supported. Note: Digits begin from left (least) to right (greatest). e.g. 1000 would be '1', 0100 would be '2'. 0010 would be '4'. How to perceive the hypotheses' bits in order: ( th <-> F. ), ( ta <-> F. ) Would be input value X's first bit, and input value Y's first bit. ( et <-> F. ), ( ze <-> F. ) would be input value X's second bit, and input value Y's second bit. In mdandysum2p2e4, one might imagine what jth or jta could be then do the math with their truths. Also limited to the restriction jth, jta are having opposite truths equivalent to the stated truth constants. (Contributed by Jarvin Udandy, 6-Sep-2016.)
⊢ (jth ↔ ⊥)    &   ⊢ (jta ↔ ⊤)    &   ⊢ (𝜑 ↔ (𝜃 ∧ 𝜏))    &   ⊢ (𝜓 ↔ (𝜂 ∧ 𝜁))    &   ⊢ (𝜒 ↔ (𝜎 ∧ 𝜌))    &   ⊢ (𝜃 ↔ jth)    &   ⊢ (𝜏 ↔ jth)    &   ⊢ (𝜂 ↔ jta)    &   ⊢ (𝜁 ↔ jta)    &   ⊢ (𝜎 ↔ jth)    &   ⊢ (𝜌 ↔ jth)    &   ⊢ (𝜇 ↔ jth)    &   ⊢ (𝜆 ↔ jth)    &   ⊢ (𝜅 ↔ ((𝜃 ⊻ 𝜏) ⊻ (𝜃 ∧ 𝜏)))    &   ⊢ (jph ↔ ((𝜂 ⊻ 𝜁) ∨ 𝜑))    &   ⊢ (jps ↔ ((𝜎 ⊻ 𝜌) ∨ 𝜓))    &   ⊢ (jch ↔ ((𝜇 ⊻ 𝜆) ∨ 𝜒))    ⇒   ⊢ ((((((((((((((((𝜑 ↔ (𝜃 ∧ 𝜏)) ∧ (𝜓 ↔ (𝜂 ∧ 𝜁))) ∧ (𝜒 ↔ (𝜎 ∧ 𝜌))) ∧ (𝜃 ↔ ⊥)) ∧ (𝜏 ↔ ⊥)) ∧ (𝜂 ↔ ⊤)) ∧ (𝜁 ↔ ⊤)) ∧ (𝜎 ↔ ⊥)) ∧ (𝜌 ↔ ⊥)) ∧ (𝜇 ↔ ⊥)) ∧ (𝜆 ↔ ⊥)) ∧ (𝜅 ↔ ((𝜃 ⊻ 𝜏) ⊻ (𝜃 ∧ 𝜏)))) ∧ (jph ↔ ((𝜂 ⊻ 𝜁) ∨ 𝜑))) ∧ (jps ↔ ((𝜎 ⊻ 𝜌) ∨ 𝜓))) ∧ (jch ↔ ((𝜇 ⊻ 𝜆) ∨ 𝜒))) → ((((𝜅 ↔ ⊥) ∧ (jph ↔ ⊥)) ∧ (jps ↔ ⊤)) ∧ (jch ↔ ⊥)))
20.36  Mathbox for Alexander van der Vekens
20.36.1  General auxiliary theorems (1)
20.36.1.1  The empty set - extension
Theorem	raaan2 42094*	Rearrange restricted quantifiers with two different restricting classes, analogous to raaan 4303. It is necessary that either both restricting classes are empty or both are not empty. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
⊢ Ⅎ𝑦𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ ((𝐴 = ∅ ↔ 𝐵 = ∅) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
20.36.1.2  Unordered and ordered pairs - extension for singletons
Theorem	eusnsn 42095*	There is a unique element of a singleton which is equal to another singleton. (Contributed by AV, 24-Aug-2022.)
⊢ ∃!𝑥{𝑥} = {𝑦}
Theorem	absnsb 42096*	If the class abstraction {𝑥 ∣ 𝜑} associated with the wff 𝜑 is a singleton, the wff is true for the singleton element. (Contributed by AV, 24-Aug-2022.)
⊢ ({𝑥 ∣ 𝜑} = {𝑦} → [𝑦 / 𝑥]𝜑)
Theorem	euabsneu 42097*	Another way to express existential uniqueness of a wff 𝜑: its associated class abstraction {𝑥 ∣ 𝜑} is a singleton. Variant of euabsn2 4492 using existential uniqueness for the singleton element instead of existence only. (Contributed by AV, 24-Aug-2022.)
⊢ (∃!𝑥𝜑 ↔ ∃!𝑦{𝑥 ∣ 𝜑} = {𝑦})
20.36.1.3  Unordered and ordered pairs - extension for unordered pairs
Theorem	elprneb 42098	An element of a proper unordered pair is the first element iff it is not the second element. (Contributed by AV, 18-Jun-2020.)
⊢ ((𝐴 ∈ {𝐵, 𝐶} ∧ 𝐵 ≠ 𝐶) → (𝐴 = 𝐵 ↔ 𝐴 ≠ 𝐶))
20.36.1.4  Relations - extension
Theorem	eubrv 42099*	If there is a unique set which is related to a class, then the class must be a set. (Contributed by AV, 25-Aug-2022.)
⊢ (∃!𝑏 𝐴𝑅𝑏 → 𝐴 ∈ V)
Theorem	eubrdm 42100*	If there is a unique set which is related to a class, then the class is an element of the domain of the relation. (Contributed by AV, 25-Aug-2022.)
⊢ (∃!𝑏 𝐴𝑅𝑏 → 𝐴 ∈ dom 𝑅)