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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | singoutnword 46501 | Singleton with character out of range 𝑉 is not a word for that range. (Contributed by Ender Ting, 21-Nov-2024.) |
⊢ 𝐴 ∈ V ⇒ ⊢ (¬ 𝐴 ∈ 𝑉 → ¬ 〈“𝐴”〉 ∈ Word 𝑉) | ||
Theorem | singoutnupword 46502 | Singleton with character out of range 𝑆 is not an increasing sequence for that range. (Contributed by Ender Ting, 22-Nov-2024.) |
⊢ 𝐴 ∈ V ⇒ ⊢ (¬ 𝐴 ∈ 𝑆 → ¬ 〈“𝐴”〉 ∈ UpWord 𝑆) | ||
Theorem | upwordsing 46503 | Singleton is an increasing sequence for any compatible range. (Contributed by Ender Ting, 21-Nov-2024.) |
⊢ 𝐴 ∈ 𝑆 ⇒ ⊢ 〈“𝐴”〉 ∈ UpWord 𝑆 | ||
Theorem | upwordsseti 46504 | Strictly increasing sequences with a set given for range form a set. (Contributed by Ender Ting, 21-Nov-2024.) |
⊢ 𝑆 ∈ V ⇒ ⊢ UpWord 𝑆 ∈ V | ||
Theorem | tworepnotupword 46505 | Concatenation of identical singletons is never an increasing sequence. (Contributed by Ender Ting, 22-Nov-2024.) |
⊢ 𝐴 ∈ V ⇒ ⊢ ¬ (〈“𝐴”〉 ++ 〈“𝐴”〉) ∈ UpWord 𝑆 | ||
Theorem | upwrdfi 46506* | 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) | ||
Theorem | hirstL-ax3 46507 | 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 46508 | Recover ax-3 8 from hirstL-ax3 46507. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.) |
⊢ ((¬ 𝜑 → ¬ 𝜓) → (𝜓 → 𝜑)) | ||
Theorem | aibandbiaiffaiffb 46509 | 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 46510 | 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 46511 | 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 46512 | Given a is equivalent to ⊤, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.) |
⊢ (𝜑 ↔ ⊤) ⇒ ⊢ 𝜑 | ||
Theorem | aisfina 46513 | Given a is equivalent to ⊥, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.) |
⊢ (𝜑 ↔ ⊥) ⇒ ⊢ ¬ 𝜑 | ||
Theorem | bothtbothsame 46514 | 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 46515 | 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 46516 | 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 46517 | 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 46518 | Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1506 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.) |
⊢ (𝜑 ⊻ 𝜓) ⇒ ⊢ ¬ (𝜑 ↔ 𝜓) | ||
Theorem | aiffnbandciffatnotciffb 46519 | 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 46520 | 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 46521 | 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 46522 | 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 46523 | 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 46524 | 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 46525 | 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 46526 | 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 46527 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.) |
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
Theorem | bisaiaisb 46528 | Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓)) | ||
Theorem | atbiffatnnbalt 46529 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.) |
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
Theorem | abnotbtaxb 46530 | Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ 𝜑 & ⊢ ¬ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
Theorem | abnotataxb 46531 | Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ ¬ 𝜑 & ⊢ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
Theorem | conimpf 46532 | 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 46533 | 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 46534 | 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 46535 | 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 46536 | 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 46537 | 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 46538 | 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 46539 | 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 46540 | 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 46541 | 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 46542 | 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 46543 | 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 46544 | 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 46545 | 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 46546 | 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 46547 | For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.) |
⊢ (𝜑 ∧ 𝜓) ⇒ ⊢ (𝜑 → 𝜓) | ||
Theorem | onenotinotbothi 46548 | 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 46549 | 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 46550 | 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 46551 | 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 46552 | 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 46553 | 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 46554 | 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 46555 | 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 46556 | 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 46557 | 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 46558 | 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 46559 | 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 46560 | 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 46561 | 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 46562 | 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 46563 | 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 46564 | 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 46565 | 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 46566 | 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 46567 | 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 46568 | 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 46569 | 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 46570 | 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 46571 | 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 46572 | 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 46573 | 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 46574 | 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 46575 | 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 46576 | 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 46577 | 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 46578 | 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 46579 | 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 46580 | 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 46581 | 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 46582 | 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 46583 | 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 46584 | 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 46585 | 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 46586 | 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 46587 | 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 46588 | 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 46589 | 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 46590 | 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 46591 | 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 46592 | 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 46593 | 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 46594 | 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 46595 | 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 46596 | 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 46597 | 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 46598 | 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 46599 | 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 46600 | 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)) |
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