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Theorem List for Metamath Proof Explorer - 44401-44500   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremaibnbna 44401 Given a implies b, (not b), there exists a proof for (not a). (Contributed by Jarvin Udandy, 1-Sep-2016.)
(𝜑𝜓)    &    ¬ 𝜓        ¬ 𝜑
 
Theoremaibnbaif 44402 Given a implies b, not b, there exists a proof for a is F. (Contributed by Jarvin Udandy, 1-Sep-2016.)
(𝜑𝜓)    &    ¬ 𝜓       (𝜑 ↔ ⊥)
 
Theoremaiffbtbat 44403 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.)
(𝜑𝜓)    &   (⊤ ↔ 𝜓)       (𝜑 ↔ ⊤)
 
Theoremastbstanbst 44404 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 ↔ ⊤)       ((𝜑𝜓) ↔ ⊤)
 
Theoremaistbistaandb 44405 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 ↔ ⊤)       (𝜑𝜓)
 
Theoremaisbnaxb 44406 Given a is equivalent to b, there exists a proof for (not (a xor b)). (Contributed by Jarvin Udandy, 28-Aug-2016.)
(𝜑𝜓)        ¬ (𝜑𝜓)
 
Theorematbiffatnnb 44407 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
((𝜑𝜓) → (𝜑 → ¬ ¬ 𝜓))
 
Theorembisaiaisb 44408 Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
((𝜓𝜑) → (𝜑𝜓))
 
Theorematbiffatnnbalt 44409 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
((𝜑𝜓) → (𝜑 → ¬ ¬ 𝜓))
 
Theoremabnotbtaxb 44410 Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
𝜑    &    ¬ 𝜓       (𝜑𝜓)
 
Theoremabnotataxb 44411 Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
¬ 𝜑    &   𝜓       (𝜑𝜓)
 
Theoremconimpf 44412 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    &   (𝜑𝜓)       (𝜑 ↔ ⊥)
 
Theoremconimpfalt 44413 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    &   (𝜑𝜓)       (𝜑 ↔ ⊥)
 
Theoremaistbisfiaxb 44414 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 ↔ ⊥)       (𝜑𝜓)
 
Theoremaisfbistiaxb 44415 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 ↔ ⊤)       (𝜑𝜓)
 
Theoremaifftbifffaibif 44416 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 ↔ ⊥)       ((𝜑𝜓) ↔ ⊥)
 
Theoremaifftbifffaibifff 44417 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 ↔ ⊥)       ((𝜑𝜓) ↔ ⊥)
 
Theorematnaiana 44418 Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
𝜑        ¬ (𝜑 → (𝜑 ∧ ¬ 𝜑))
 
Theoremainaiaandna 44419 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)))
 
Theoremabcdta 44420 Given (((a and b) and c) and d), there exists a proof for a. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜑
 
Theoremabcdtb 44421 Given (((a and b) and c) and d), there exists a proof for b. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜓
 
Theoremabcdtc 44422 Given (((a and b) and c) and d), there exists a proof for c. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜒
 
Theoremabcdtd 44423 Given (((a and b) and c) and d), there exists a proof for d. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((𝜑𝜓) ∧ 𝜒) ∧ 𝜃)       𝜃
 
Theoremabciffcbatnabciffncba 44424 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. Closed form. (Contributed by Jarvin Udandy, 7-Sep-2020.)
(¬ ((𝜑𝜓) ∧ 𝜒) → ¬ ((𝜒𝜓) ∧ 𝜑))
 
Theoremabciffcbatnabciffncbai 44425 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)
(((𝜑𝜓) ∧ 𝜒) ↔ ((𝜒𝜓) ∧ 𝜑))       (¬ ((𝜑𝜓) ∧ 𝜒) → ¬ ((𝜒𝜓) ∧ 𝜑))
 
Theoremnabctnabc 44426 not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)
¬ (𝜑 → (𝜓𝜒))       𝜑 → (𝜓𝜒))
 
Theoremjabtaib 44427 For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
(𝜑𝜓)       (𝜑𝜓)
 
Theoremonenotinotbothi 44428 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.)
¬ (𝜑𝜓)        ¬ ((𝜑𝜓) ∧ (𝜒𝜃))
 
Theoremtwonotinotbothi 44429 From these two negated implications it is not the case their nonnegated forms are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
¬ (𝜑𝜓)    &    ¬ (𝜒𝜃)        ¬ ((𝜑𝜓) ∧ (𝜒𝜃))
 
Theoremclifte 44430 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) ∨ (𝜓𝜒)))
 
Theoremcliftet 44431 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
(𝜑𝜒)    &   𝜃       (𝜃 ↔ ((𝜑𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))
 
Theoremclifteta 44432 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
((𝜑 ∧ ¬ 𝜒) ∨ (𝜓𝜒))    &   𝜃       (𝜃 ↔ ((𝜑 ∧ ¬ 𝜒) ∨ (𝜓𝜒)))
 
Theoremcliftetb 44433 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
((𝜑𝜒) ∨ (𝜓 ∧ ¬ 𝜒))    &   𝜃       (𝜃 ↔ ((𝜑𝜒) ∨ (𝜓 ∧ ¬ 𝜒)))
 
Theoremconfun 44434 Given the hypotheses there exists a proof for (c implies ( d iff a ) ). (Contributed by Jarvin Udandy, 6-Sep-2020.)
𝜑    &   (𝜒𝜓)    &   (𝜒𝜃)    &   (𝜑 → (𝜑𝜓))       (𝜒 → (𝜃𝜑))
 
Theoremconfun2 44435 Confun simplified to two propositions. (Contributed by Jarvin Udandy, 6-Sep-2020.)
(𝜓𝜑)    &   (𝜓 → ¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)))    &   ((𝜓𝜑) → ((𝜓𝜑) → 𝜑))       (𝜓 → (¬ (𝜓 → (𝜓 ∧ ¬ 𝜓)) ↔ (𝜓𝜑)))
 
Theoremconfun3 44436 Confun's more complex form where both a,d have been "defined". (Contributed by Jarvin Udandy, 6-Sep-2020.)
(𝜑 ↔ (𝜒𝜓))    &   (𝜃 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   (𝜒𝜓)    &   (𝜒 → ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   ((𝜒𝜓) → ((𝜒𝜓) → 𝜓))       (𝜒 → (¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)) ↔ (𝜒𝜓)))
 
Theoremconfun4 44437 An attempt at derivative. Resisted simplest path to a proof. (Contributed by Jarvin Udandy, 6-Sep-2020.)
𝜑    &   ((𝜑𝜓) → 𝜓)    &   (𝜓 → (𝜑𝜒))    &   ((𝜒𝜃) → ((𝜑𝜃) ↔ 𝜓))    &   (𝜏 ↔ (𝜒𝜃))    &   (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   𝜓    &   (𝜒𝜃)       (𝜒 → (𝜓𝜏))
 
Theoremconfun5 44438 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    &   ((𝜑𝜓) → 𝜓)    &   (𝜓 → (𝜑𝜒))    &   ((𝜒𝜃) → ((𝜑𝜃) ↔ 𝜓))    &   (𝜏 ↔ (𝜒𝜃))    &   (𝜂 ↔ ¬ (𝜒 → (𝜒 ∧ ¬ 𝜒)))    &   𝜓    &   (𝜒𝜃)       (𝜒 → (𝜂𝜏))
 
Theoremplcofph 44439 Given, a,b and a "definition" for c, c is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜒 ↔ ((((𝜑𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   𝜑    &   𝜓       𝜒
 
Theorempldofph 44440 Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜏 ↔ ((𝜒𝜃) ∧ (𝜑𝜒) ∧ ((𝜑𝜓) → (𝜓𝜃))))    &   𝜑    &   𝜓    &   𝜒    &   𝜃       𝜏
 
Theoremplvcofph 44441 Given, a,b,d, and "definitions" for c, e, f: f is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜒 ↔ ((((𝜑𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   (𝜏 ↔ ((𝜒𝜃) ∧ (𝜑𝜒) ∧ ((𝜑𝜓) → (𝜓𝜃))))    &   (𝜂 ↔ (𝜒𝜏))    &   𝜑    &   𝜓    &   𝜃       𝜂
 
Theoremplvcofphax 44442 Given, a,b,d, and "definitions" for c, e, f, g: g is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜒 ↔ ((((𝜑𝜓) ↔ 𝜑) → (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))) ∧ (𝜑 ∧ ¬ (𝜑 ∧ ¬ 𝜑))))    &   (𝜏 ↔ ((𝜒𝜃) ∧ (𝜑𝜒) ∧ ((𝜑𝜓) → (𝜓𝜃))))    &   (𝜂 ↔ (𝜒𝜏))    &   𝜑    &   𝜓    &   𝜃    &   (𝜁 ↔ ¬ (𝜓 ∧ ¬ 𝜏))       𝜁
 
Theoremplvofpos 44443 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 ↔ (¬ 𝜑𝜓))    &   (𝜏 ↔ (𝜑 ∧ ¬ 𝜓))    &   (𝜂 ↔ (𝜑𝜓))    &   (𝜁 ↔ (((((¬ ((𝜇𝜒) ∧ (𝜇𝜃)) ∧ ¬ ((𝜇𝜒) ∧ (𝜇𝜏))) ∧ ¬ ((𝜇𝜒) ∧ (𝜒𝜂))) ∧ ¬ ((𝜇𝜃) ∧ (𝜇𝜏))) ∧ ¬ ((𝜇𝜃) ∧ (𝜇𝜂))) ∧ ¬ ((𝜇𝜏) ∧ (𝜇𝜂))))    &   (𝜎 ↔ (((𝜇𝜒) ∨ (𝜇𝜃)) ∨ ((𝜇𝜏) ∨ (𝜇𝜂))))    &   (𝜌 ↔ (𝜁𝜎))    &   𝜁    &   𝜎       𝜌
 
Theoremmdandyv0 44444 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))
 
Theoremmdandyv1 44445 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))
 
Theoremmdandyv2 44446 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))
 
Theoremmdandyv3 44447 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜑))
 
Theoremmdandyv4 44448 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))
 
Theoremmdandyv5 44449 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))
 
Theoremmdandyv6 44450 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 ↔ ⊥)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))
 
Theoremmdandyv7 44451 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 ↔ ⊥)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜑))
 
Theoremmdandyv8 44452 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))
 
Theoremmdandyv9 44453 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))
 
Theoremmdandyv10 44454 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))
 
Theoremmdandyv11 44455 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜑)) ∧ (𝜂𝜓))
 
Theoremmdandyv12 44456 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))
 
Theoremmdandyv13 44457 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜑)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))
 
Theoremmdandyv14 44458 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 ↔ ⊤)       ((((𝜒𝜑) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))
 
Theoremmdandyv15 44459 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 ↔ ⊤)       ((((𝜒𝜓) ∧ (𝜃𝜓)) ∧ (𝜏𝜓)) ∧ (𝜂𝜓))
 
Theoremmdandyvr0 44460 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvr1 44461 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvr2 44462 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvr3 44463 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvr4 44464 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvr5 44465 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvr6 44466 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvr7 44467 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvr8 44468 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvr9 44469 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvr10 44470 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvr11 44471 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvr12 44472 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvr13 44473 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvr14 44474 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvr15 44475 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx0 44476 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx1 44477 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx2 44478 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx3 44479 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx4 44480 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx5 44481 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx6 44482 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx7 44483 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜑)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜁))
 
Theoremmdandyvrx8 44484 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx9 44485 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx10 44486 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx11 44487 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜑)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜁)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx12 44488 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx13 44489 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜑)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜁)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx14 44490 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜑)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜁) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
Theoremmdandyvrx15 44491 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.)
(𝜑𝜁)    &   (𝜓𝜎)    &   (𝜒𝜓)    &   (𝜃𝜓)    &   (𝜏𝜓)    &   (𝜂𝜓)       ((((𝜒𝜎) ∧ (𝜃𝜎)) ∧ (𝜏𝜎)) ∧ (𝜂𝜎))
 
TheoremH15NH16TH15IH16 44492 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       (((((((((((((((𝜑𝜓) ∧ 𝜒) ∧ 𝜃) ∧ 𝜏) ∧ 𝜂) ∧ 𝜁) ∧ 𝜎) ∧ 𝜌) ∧ 𝜇) ∧ 𝜆) ∧ 𝜅) ∧ jph) ∧ jps) ∧ jch) → jth)
 
Theoremdandysum2p2e4 44493 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 ↔ (𝜃𝜏))    &   (𝜓 ↔ (𝜂𝜁))    &   (𝜒 ↔ (𝜎𝜌))    &   (𝜃 ↔ ⊥)    &   (𝜏 ↔ ⊥)    &   (𝜂 ↔ ⊤)    &   (𝜁 ↔ ⊤)    &   (𝜎 ↔ ⊥)    &   (𝜌 ↔ ⊥)    &   (𝜇 ↔ ⊥)    &   (𝜆 ↔ ⊥)    &   (𝜅 ↔ ((𝜃𝜏) ⊻ (𝜃𝜏)))    &   (jph ↔ ((𝜂𝜁) ∨ 𝜑))    &   (jps ↔ ((𝜎𝜌) ∨ 𝜓))    &   (jch ↔ ((𝜇𝜆) ∨ 𝜒))       ((((((((((((((((𝜑 ↔ (𝜃𝜏)) ∧ (𝜓 ↔ (𝜂𝜁))) ∧ (𝜒 ↔ (𝜎𝜌))) ∧ (𝜃 ↔ ⊥)) ∧ (𝜏 ↔ ⊥)) ∧ (𝜂 ↔ ⊤)) ∧ (𝜁 ↔ ⊤)) ∧ (𝜎 ↔ ⊥)) ∧ (𝜌 ↔ ⊥)) ∧ (𝜇 ↔ ⊥)) ∧ (𝜆 ↔ ⊥)) ∧ (𝜅 ↔ ((𝜃𝜏) ⊻ (𝜃𝜏)))) ∧ (jph ↔ ((𝜂𝜁) ∨ 𝜑))) ∧ (jps ↔ ((𝜎𝜌) ∨ 𝜓))) ∧ (jch ↔ ((𝜇𝜆) ∨ 𝜒))) → ((((𝜅 ↔ ⊥) ∧ (jph ↔ ⊥)) ∧ (jps ↔ ⊤)) ∧ (jch ↔ ⊥)))
 
Theoremmdandysum2p2e4 44494 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 ↔ (𝜃𝜏))    &   (𝜓 ↔ (𝜂𝜁))    &   (𝜒 ↔ (𝜎𝜌))    &   (𝜃jth)    &   (𝜏jth)    &   (𝜂jta)    &   (𝜁jta)    &   (𝜎jth)    &   (𝜌jth)    &   (𝜇jth)    &   (𝜆jth)    &   (𝜅 ↔ ((𝜃𝜏) ⊻ (𝜃𝜏)))    &   (jph ↔ ((𝜂𝜁) ∨ 𝜑))    &   (jps ↔ ((𝜎𝜌) ∨ 𝜓))    &   (jch ↔ ((𝜇𝜆) ∨ 𝜒))       ((((((((((((((((𝜑 ↔ (𝜃𝜏)) ∧ (𝜓 ↔ (𝜂𝜁))) ∧ (𝜒 ↔ (𝜎𝜌))) ∧ (𝜃 ↔ ⊥)) ∧ (𝜏 ↔ ⊥)) ∧ (𝜂 ↔ ⊤)) ∧ (𝜁 ↔ ⊤)) ∧ (𝜎 ↔ ⊥)) ∧ (𝜌 ↔ ⊥)) ∧ (𝜇 ↔ ⊥)) ∧ (𝜆 ↔ ⊥)) ∧ (𝜅 ↔ ((𝜃𝜏) ⊻ (𝜃𝜏)))) ∧ (jph ↔ ((𝜂𝜁) ∨ 𝜑))) ∧ (jps ↔ ((𝜎𝜌) ∨ 𝜓))) ∧ (jch ↔ ((𝜇𝜆) ∨ 𝜒))) → ((((𝜅 ↔ ⊥) ∧ (jph ↔ ⊥)) ∧ (jps ↔ ⊤)) ∧ (jch ↔ ⊥)))
 
20.40  Mathbox for Adhemar
 
Theoremadh-jarrsc 44495 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.)
(((𝜑𝜓) → 𝜒) → (𝜃 → (𝜓𝜒)))
 
20.40.1  Minimal implicational calculus

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 83, ax-1 6, pm2.43 56 } (written B'KW), or with { imim2 58, pm2.04 90, ax-1 6, pm2.43 56 } (written BCKW), or with the single axiom adh-minim 44496, or with the single axiom adh-minimp 44508. This section proves first adh-minim 44496 from { ax-1 6, ax-2 7 }, followed by the converse, due to Ivo Thomas; and then it proves adh-minimp 44508 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 44496 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 44496 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.)

 
Theoremadh-minim 44496 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.)
(((𝜑𝜓) → 𝜒) → (𝜃 → ((𝜓 → (𝜒𝜏)) → (𝜓𝜏))))
 
Theoremadh-minim-ax1-ax2-lem1 44497 First lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 44496 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 → (𝜓𝜏)) → (𝜃𝜏))) → 𝜂)) → (𝜓𝜂)))
 
Theoremadh-minim-ax1-ax2-lem2 44498 Second lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 44496 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 → (𝜑𝜃)) → (𝜒𝜃))) → 𝜏)) → (𝜑𝜏))
 
Theoremadh-minim-ax1-ax2-lem3 44499 Third lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 44496 and ax-mp 5. Polish prefix notation: CCpCqrCqCsCpr . (Contributed by ADH, 10-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝜑 → (𝜓𝜒)) → (𝜓 → (𝜃 → (𝜑𝜒))))
 
Theoremadh-minim-ax1-ax2-lem4 44500 Fourth lemma for the derivation of ax-1 6 and ax-2 7 from adh-minim 44496 and ax-mp 5. Polish prefix notation: CCCpqrCCqCrsCqs . (Contributed by ADH, 10-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.)
(((𝜑𝜓) → 𝜒) → ((𝜓 → (𝜒𝜃)) → (𝜓𝜃)))
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