![]() |
Metamath
Proof Explorer Theorem List (p. 457 of 479) | < Previous Next > |
Bad symbols? Try the
GIF version. |
||
Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List |
Color key: | ![]() (1-30166) |
![]() (30167-31689) |
![]() (31690-47842) |
Type | Label | Description |
---|---|---|
Statement | ||
Theorem | aiffbbtat 45601 | 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 45602 | 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 45603 | Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1510 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.) |
⊢ (𝜑 ⊻ 𝜓) ⇒ ⊢ ¬ (𝜑 ↔ 𝜓) | ||
Theorem | aiffnbandciffatnotciffb 45604 | 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 45605 | 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 45606 | 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 45607 | 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 45608 | 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 45609 | 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 45610 | 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 45611 | 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 45612 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.) |
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
Theorem | bisaiaisb 45613 | Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓)) | ||
Theorem | atbiffatnnbalt 45614 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.) |
⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
Theorem | abnotbtaxb 45615 | Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ 𝜑 & ⊢ ¬ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
Theorem | abnotataxb 45616 | Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
⊢ ¬ 𝜑 & ⊢ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
Theorem | conimpf 45617 | 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 45618 | 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 45619 | 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 45620 | 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 45621 | 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 45622 | 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 45623 | 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 45624 | 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 45625 | 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 45626 | 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 45627 | 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 45628 | 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 45629 | 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 45630 | 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 45631 | 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 45632 | For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.) |
⊢ (𝜑 ∧ 𝜓) ⇒ ⊢ (𝜑 → 𝜓) | ||
Theorem | onenotinotbothi 45633 | 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 45634 | 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 45635 | 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 45636 | 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 45637 | 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 45638 | 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 45639 | 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 45640 | 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 45641 | 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 45642 | 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 45643 | 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 45644 | 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 45645 | 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 45646 | 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 45647 | 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 45648 | 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 45649 | 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 45650 | 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 45651 | 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 45652 | 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 45653 | 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 45654 | 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 45655 | 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 45656 | 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 45657 | 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 45658 | 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 45659 | 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 45660 | 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 45661 | 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 45662 | 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 45663 | 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 45664 | 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 45665 | 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 45666 | 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 45667 | 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 45668 | 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 45669 | 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 45670 | 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 45671 | 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 45672 | 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 45673 | 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 45674 | 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 45675 | 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 45676 | 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 45677 | 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 45678 | 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 45679 | 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 45680 | 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 45681 | 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 45682 | 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 45683 | 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 45684 | 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 45685 | 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 45686 | 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 45687 | 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 45688 | 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 45689 | 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 45690 | 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 45691 | 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 45692 | 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 45693 | 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 45694 | 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 45695 | 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 45696 | 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 45697 | 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 45698 |
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 would 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 45699 |
CONTRADICTION PROVED AT 1 + 1 = 2 . Luckily Mario Carneiro did a
successful version of his own.
See Mario's Relevant Work: 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 would 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 ↔ ⊥))) | ||
Theorem | adh-jarrsc 45700 | Replacement of a nested antecedent with an outer antecedent. Commuted simplificated form of elimination of a nested antecedent. Also holds intuitionistically. Polish prefix notation: CCCpqrCsCqr . (Contributed by ADH, 10-Nov-2023.) (Proof modification is discouraged.) |
⊢ (((𝜑 → 𝜓) → 𝜒) → (𝜃 → (𝜓 → 𝜒))) |
< Previous Next > |
Copyright terms: Public domain | < Previous Next > |