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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | axorbciffatcxorb 47801 | 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 47802 | 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 47803 | 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 47804 | 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 47805 | 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 47806 | 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 47807 | 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 47808 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.) |
| ⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
| Theorem | bisaiaisb 47809 | Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.) |
| ⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓)) | ||
| Theorem | atbiffatnnbalt 47810 | If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.) |
| ⊢ ((𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓)) | ||
| Theorem | abnotbtaxb 47811 | Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
| ⊢ 𝜑 & ⊢ ¬ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
| Theorem | abnotataxb 47812 | Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.) |
| ⊢ ¬ 𝜑 & ⊢ 𝜓 ⇒ ⊢ (𝜑 ⊻ 𝜓) | ||
| Theorem | conimpf 47813 | 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 47814 | 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 47815 | 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 47816 | 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 47817 | 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 47818 | 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 47819 | 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 47820 | 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 47821 | 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 47822 | 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 47823 | 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 47824 | 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 47825 | 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 47826 | 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 47827 | 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 47828 | For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.) |
| ⊢ (𝜑 ∧ 𝜓) ⇒ ⊢ (𝜑 → 𝜓) | ||
| Theorem | onenotinotbothi 47829 | 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 47830 | 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 47831 | 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 47832 | 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 47833 | 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 47834 | 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 47835 | 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 47836 | 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 47837 | 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 47838 | 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 47839 | 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 47840 | 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 47841 | 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 47842 | 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 47843 | 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 47844 | 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 47845 | 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 47846 | 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 47847 | 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 47848 | 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 47849 | 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 47850 | 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 47851 | 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 47852 | 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 47853 | 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 47854 | 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 47855 | 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 47856 | 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 47857 | 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 47858 | 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 47859 | 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 47860 | 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 47861 | 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 47862 | 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 47863 | 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 47864 | 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 47865 | 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 47866 | 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 47867 | 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 47868 | 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 47869 | 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 47870 | 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 47871 | 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 47872 | 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 47873 | 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 47874 | 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 47875 | 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 47876 | 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 47877 | 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 47878 | 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 47879 | 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 47880 | 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 47881 | 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 47882 | 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 47883 | 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 47884 | 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 47885 | 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 47886 | 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 47887 | 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 47888 | 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 47889 | 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 47890 | 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 47891 | 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 47892 | 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 47893 | 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 47894 |
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 47895 |
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 47896 | 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))) | ||
Minimal implicational calculus, or intuitionistic implicational calculus, or positive implicational calculus, is the implicational fragment of minimal calculus (which is also the implicational fragment of intuitionistic calculus and of positive calculus). It is sometimes called "C-pure intuitionism" since the letter C is used to denote implication in Polish prefix notation. It can be axiomatized by the inference rule of modus ponens ax-mp 5 together with the axioms { ax-1 6, ax-2 7 } (sometimes written KS), or with { imim1 84, ax-1 6, pm2.43 57 } (written B'KW), or with { imim2 59, pm2.04 91, ax-1 6, pm2.43 57 } (written BCKW), or with the single axiom adh-minim 47897, or with the single axiom adh-minimp 47909. This section proves first adh-minim 47897 from { ax-1 6, ax-2 7 }, followed by the converse, due to Ivo Thomas; and then it proves adh-minimp 47909 from { ax-1 6, ax-2 7 }, also followed by the converse, also due to Ivo Thomas. Sources for this section are * Carew Arthur Meredith, A single axiom of positive logic, The Journal of Computing Systems, volume 1, issue 3, July 1953, pages 169--170; * Ivo Thomas, On Meredith's sole positive axiom, Notre Dame Journal of Formal Logic, volume XV, number 3, July 1974, page 477, in which the derivations of { ax-1 6, ax-2 7 } from adh-minim 47897 are shortened (compared to Meredith's derivations in the aforementioned paper); * Carew Arthur Meredith and Arthur Norman Prior, Notes on the axiomatics of the propositional calculus, Notre Dame Journal of Formal Logic, volume IV, number 3, July 1963, pages 171--187; and * the webpage https://web.ics.purdue.edu/~dulrich/C-pure-intuitionism-page.htm 47897 on Dolph Edward "Ted" Ulrich's website, where these and other single axioms for the minimal implicational calculus are listed. This entire section also holds intuitionistically. Users of the Polish prefix notation also often use a compact notation for proof derivations known as the D-notation where "D" stands for "condensed Detachment". For instance, "D21" means detaching ax-1 6 from ax-2 7, that is, using modus ponens ax-mp 5 with ax-1 6 as minor premise and ax-2 7 as major premise. When the numbered lemmas surpass 10, dots are added between the numbers. D-strings are accepted by the grammar Dundotted := digit | "D" Dundotted Dundotted ; Ddotted := digit + | "D" Ddotted "." Ddotted ; Dstr := Dundotted | Ddotted . (Contributed by BJ, 11-Apr-2021.) (Revised by ADH, 10-Nov-2023.) | ||
| Theorem | adh-minim 47897 | A single axiom for minimal implicational calculus, due to Meredith. Other single axioms of the same length are known, but it is thought to be the minimal length. This is the axiom from Carew Arthur Meredith, A single axiom of positive logic, The Journal of Computing Systems, volume 1, issue 3, July 1953, pages 169--170. A two-line review by Alonzo Church of this article can be found in The Journal of Symbolic Logic, volume 19, issue 2, June 1954, page 144, https://doi.org/10.2307/2268914. Known as "HI-1" on Dolph Edward "Ted" Ulrich's web page. In the next 6 lemmas and 3 theorems, ax-1 6 and ax-2 7 are derived from this single axiom in 16 detachments (instances of ax-mp 5) in total. Polish prefix notation: CCCpqrCsCCqCrtCqt . (Contributed by ADH, 10-Nov-2023.) |
| ⊢ (((𝜑 → 𝜓) → 𝜒) → (𝜃 → ((𝜓 → (𝜒 → 𝜏)) → (𝜓 → 𝜏)))) | ||
| Theorem | adh-minim-ax1-ax2-lem1 47898 | First lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 47897 and ax-mp 5. Polish prefix notation: CpCCqCCrCCsCqtCstuCqu . (Contributed by ADH, 10-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ (𝜑 → ((𝜓 → ((𝜒 → ((𝜃 → (𝜓 → 𝜏)) → (𝜃 → 𝜏))) → 𝜂)) → (𝜓 → 𝜂))) | ||
| Theorem | adh-minim-ax1-ax2-lem2 47899 | Second lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 47897 and ax-mp 5. Polish prefix notation: CCpCCqCCrCpsCrstCpt . (Contributed by ADH, 10-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ ((𝜑 → ((𝜓 → ((𝜒 → (𝜑 → 𝜃)) → (𝜒 → 𝜃))) → 𝜏)) → (𝜑 → 𝜏)) | ||
| Theorem | adh-minim-ax1-ax2-lem3 47900 | Third lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 47897 and ax-mp 5. Polish prefix notation: CCpCqrCqCsCpr . (Contributed by ADH, 10-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.) |
| ⊢ ((𝜑 → (𝜓 → 𝜒)) → (𝜓 → (𝜃 → (𝜑 → 𝜒)))) | ||
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