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Theorem List for Metamath Proof Explorer - 45601-45700   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremupwrdfi 45601* There is a finite number of strictly increasing sequences of a given length over finite alphabet. Trivially holds for invalid lengths where there're zero matching sequences. (Contributed by Ender Ting, 5-Jan-2024.)
(𝑆 ∈ Fin β†’ {π‘Ž ∈ UpWord 𝑆 ∣ (β™―β€˜π‘Ž) = 𝑇} ∈ Fin)
 
21.43  Mathbox for Jarvin Udandy
 
TheoremhirstL-ax3 45602 The third axiom of a system called "L" but proven to be a theorem since set.mm uses a different third axiom. This is named hirst after Holly P. Hirst and Jeffry L. Hirst. Axiom A3 of [Mendelson] p. 35. (Contributed by Jarvin Udandy, 7-Feb-2015.) (Proof modification is discouraged.)
((Β¬ πœ‘ β†’ Β¬ πœ“) β†’ ((Β¬ πœ‘ β†’ πœ“) β†’ πœ‘))
 
Theoremax3h 45603 Recover ax-3 8 from hirstL-ax3 45602. (Contributed by Jarvin Udandy, 3-Jul-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
((Β¬ πœ‘ β†’ Β¬ πœ“) β†’ (πœ“ β†’ πœ‘))
 
Theoremaibandbiaiffaiffb 45604 A closed form showing (a implies b and b implies a) same-as (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((πœ‘ β†’ πœ“) ∧ (πœ“ β†’ πœ‘)) ↔ (πœ‘ ↔ πœ“))
 
Theoremaibandbiaiaiffb 45605 A closed form showing (a implies b and b implies a) implies (a same-as b). (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((πœ‘ β†’ πœ“) ∧ (πœ“ β†’ πœ‘)) β†’ (πœ‘ ↔ πœ“))
 
Theoremnotatnand 45606 Do not use. Use intnanr instead. Given not a, there exists a proof for not (a and b). (Contributed by Jarvin Udandy, 31-Aug-2016.)
Β¬ πœ‘    β‡’    Β¬ (πœ‘ ∧ πœ“)
 
Theoremaistia 45607 Given a is equivalent to ⊀, there exists a proof for a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(πœ‘ ↔ ⊀)    β‡’   πœ‘
 
Theoremaisfina 45608 Given a is equivalent to βŠ₯, there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(πœ‘ ↔ βŠ₯)    β‡’    Β¬ πœ‘
 
Theorembothtbothsame 45609 Given both a, b are equivalent to ⊀, there exists a proof for a is the same as b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
(πœ‘ ↔ ⊀)    &   (πœ“ ↔ ⊀)    β‡’   (πœ‘ ↔ πœ“)
 
Theorembothfbothsame 45610 Given both a, b are equivalent to βŠ₯, there exists a proof for a is the same as b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ βŠ₯)    β‡’   (πœ‘ ↔ πœ“)
 
Theoremaiffbbtat 45611 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.)
(πœ‘ ↔ πœ“)    &   (πœ“ ↔ ⊀)    β‡’   (πœ‘ ↔ ⊀)
 
Theoremaisbbisfaisf 45612 Given a is equivalent to b, b is equivalent to βŠ₯ there exists a proof for a is equivalent to F. (Contributed by Jarvin Udandy, 30-Aug-2016.)
(πœ‘ ↔ πœ“)    &   (πœ“ ↔ βŠ₯)    β‡’   (πœ‘ ↔ βŠ₯)
 
Theoremaxorbtnotaiffb 45613 Given a is exclusive to b, there exists a proof for (not (a if-and-only-if b)); df-xor 1511 is a closed form of this. (Contributed by Jarvin Udandy, 7-Sep-2016.)
(πœ‘ ⊻ πœ“)    β‡’    Β¬ (πœ‘ ↔ πœ“)
 
Theoremaiffnbandciffatnotciffb 45614 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.)
(πœ‘ ↔ Β¬ πœ“)    &   (πœ’ ↔ πœ‘)    β‡’    Β¬ (πœ’ ↔ πœ“)
 
Theoremaxorbciffatcxorb 45615 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.)
(πœ‘ ⊻ πœ“)    &   (πœ’ ↔ πœ‘)    β‡’   (πœ’ ⊻ πœ“)
 
Theoremaibnbna 45616 Given a implies b, (not b), there exists a proof for (not a). (Contributed by Jarvin Udandy, 1-Sep-2016.)
(πœ‘ β†’ πœ“)    &    Β¬ πœ“    β‡’    Β¬ πœ‘
 
Theoremaibnbaif 45617 Given a implies b, not b, there exists a proof for a is F. (Contributed by Jarvin Udandy, 1-Sep-2016.)
(πœ‘ β†’ πœ“)    &    Β¬ πœ“    β‡’   (πœ‘ ↔ βŠ₯)
 
Theoremaiffbtbat 45618 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.)
(πœ‘ ↔ πœ“)    &   (⊀ ↔ πœ“)    β‡’   (πœ‘ ↔ ⊀)
 
Theoremastbstanbst 45619 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.)
(πœ‘ ↔ ⊀)    &   (πœ“ ↔ ⊀)    β‡’   ((πœ‘ ∧ πœ“) ↔ ⊀)
 
Theoremaistbistaandb 45620 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.)
(πœ‘ ↔ ⊀)    &   (πœ“ ↔ ⊀)    β‡’   (πœ‘ ∧ πœ“)
 
Theoremaisbnaxb 45621 Given a is equivalent to b, there exists a proof for (not (a xor b)). (Contributed by Jarvin Udandy, 28-Aug-2016.)
(πœ‘ ↔ πœ“)    β‡’    Β¬ (πœ‘ ⊻ πœ“)
 
Theorematbiffatnnb 45622 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 28-Aug-2016.)
((πœ‘ β†’ πœ“) β†’ (πœ‘ β†’ Β¬ Β¬ πœ“))
 
Theorembisaiaisb 45623 Application of bicom1 with a, b swapped. (Contributed by Jarvin Udandy, 31-Aug-2016.)
((πœ“ ↔ πœ‘) β†’ (πœ‘ ↔ πœ“))
 
Theorematbiffatnnbalt 45624 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
((πœ‘ β†’ πœ“) β†’ (πœ‘ β†’ Β¬ Β¬ πœ“))
 
Theoremabnotbtaxb 45625 Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
πœ‘    &    Β¬ πœ“    β‡’   (πœ‘ ⊻ πœ“)
 
Theoremabnotataxb 45626 Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
Β¬ πœ‘    &   πœ“    β‡’   (πœ‘ ⊻ πœ“)
 
Theoremconimpf 45627 Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 28-Aug-2016.)
πœ‘    &    Β¬ πœ“    &   (πœ‘ β†’ πœ“)    β‡’   (πœ‘ ↔ βŠ₯)
 
Theoremconimpfalt 45628 Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 29-Aug-2016.)
πœ‘    &    Β¬ πœ“    &   (πœ‘ β†’ πœ“)    β‡’   (πœ‘ ↔ βŠ₯)
 
Theoremaistbisfiaxb 45629 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.)
(πœ‘ ↔ ⊀)    &   (πœ“ ↔ βŠ₯)    β‡’   (πœ‘ ⊻ πœ“)
 
Theoremaisfbistiaxb 45630 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    β‡’   (πœ‘ ⊻ πœ“)
 
Theoremaifftbifffaibif 45631 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.)
(πœ‘ ↔ ⊀)    &   (πœ“ ↔ βŠ₯)    β‡’   ((πœ‘ β†’ πœ“) ↔ βŠ₯)
 
Theoremaifftbifffaibifff 45632 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.)
(πœ‘ ↔ ⊀)    &   (πœ“ ↔ βŠ₯)    β‡’   ((πœ‘ ↔ πœ“) ↔ βŠ₯)
 
Theorematnaiana 45633 Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
πœ‘    β‡’    Β¬ (πœ‘ β†’ (πœ‘ ∧ Β¬ πœ‘))
 
Theoremainaiaandna 45634 Given a, a implies it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
πœ‘    β‡’   (πœ‘ β†’ Β¬ (πœ‘ β†’ (πœ‘ ∧ Β¬ πœ‘)))
 
Theoremabcdta 45635 Given (((a and b) and c) and d), there exists a proof for a. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((πœ‘ ∧ πœ“) ∧ πœ’) ∧ πœƒ)    β‡’   πœ‘
 
Theoremabcdtb 45636 Given (((a and b) and c) and d), there exists a proof for b. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((πœ‘ ∧ πœ“) ∧ πœ’) ∧ πœƒ)    β‡’   πœ“
 
Theoremabcdtc 45637 Given (((a and b) and c) and d), there exists a proof for c. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((πœ‘ ∧ πœ“) ∧ πœ’) ∧ πœƒ)    β‡’   πœ’
 
Theoremabcdtd 45638 Given (((a and b) and c) and d), there exists a proof for d. (Contributed by Jarvin Udandy, 3-Sep-2016.)
(((πœ‘ ∧ πœ“) ∧ πœ’) ∧ πœƒ)    β‡’   πœƒ
 
Theoremabciffcbatnabciffncba 45639 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. Closed form. (Contributed by Jarvin Udandy, 7-Sep-2020.)
(Β¬ ((πœ‘ ∧ πœ“) ∧ πœ’) β†’ Β¬ ((πœ’ ∧ πœ“) ∧ πœ‘))
 
Theoremabciffcbatnabciffncbai 45640 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)
(((πœ‘ ∧ πœ“) ∧ πœ’) ↔ ((πœ’ ∧ πœ“) ∧ πœ‘))    β‡’   (Β¬ ((πœ‘ ∧ πœ“) ∧ πœ’) β†’ Β¬ ((πœ’ ∧ πœ“) ∧ πœ‘))
 
Theoremnabctnabc 45641 not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)
Β¬ (πœ‘ β†’ (πœ“ ∧ πœ’))    β‡’   (Β¬ πœ‘ β†’ (πœ“ ∧ πœ’))
 
Theoremjabtaib 45642 For when pm3.4 lacks a pm3.4i. (Contributed by Jarvin Udandy, 9-Sep-2020.)
(πœ‘ ∧ πœ“)    β‡’   (πœ‘ β†’ πœ“)
 
Theoremonenotinotbothi 45643 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.)
Β¬ (πœ‘ β†’ πœ“)    β‡’    Β¬ ((πœ‘ β†’ πœ“) ∧ (πœ’ β†’ πœƒ))
 
Theoremtwonotinotbothi 45644 From these two negated implications it is not the case their nonnegated forms are both true. (Contributed by Jarvin Udandy, 11-Sep-2020.)
Β¬ (πœ‘ β†’ πœ“)    &    Β¬ (πœ’ β†’ πœƒ)    β‡’    Β¬ ((πœ‘ β†’ πœ“) ∧ (πœ’ β†’ πœƒ))
 
Theoremclifte 45645 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
(πœ‘ ∧ Β¬ πœ’)    &   πœƒ    β‡’   (πœƒ ↔ ((πœ‘ ∧ Β¬ πœ’) ∨ (πœ“ ∧ πœ’)))
 
Theoremcliftet 45646 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
(πœ‘ ∧ πœ’)    &   πœƒ    β‡’   (πœƒ ↔ ((πœ‘ ∧ πœ’) ∨ (πœ“ ∧ Β¬ πœ’)))
 
Theoremclifteta 45647 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
((πœ‘ ∧ Β¬ πœ’) ∨ (πœ“ ∧ πœ’))    &   πœƒ    β‡’   (πœƒ ↔ ((πœ‘ ∧ Β¬ πœ’) ∨ (πœ“ ∧ πœ’)))
 
Theoremcliftetb 45648 show d is the same as an if-else involving a,b. (Contributed by Jarvin Udandy, 20-Sep-2020.)
((πœ‘ ∧ πœ’) ∨ (πœ“ ∧ Β¬ πœ’))    &   πœƒ    β‡’   (πœƒ ↔ ((πœ‘ ∧ πœ’) ∨ (πœ“ ∧ Β¬ πœ’)))
 
Theoremconfun 45649 Given the hypotheses there exists a proof for (c implies ( d iff a ) ). (Contributed by Jarvin Udandy, 6-Sep-2020.)
πœ‘    &   (πœ’ β†’ πœ“)    &   (πœ’ β†’ πœƒ)    &   (πœ‘ β†’ (πœ‘ β†’ πœ“))    β‡’   (πœ’ β†’ (πœƒ ↔ πœ‘))
 
Theoremconfun2 45650 Confun simplified to two propositions. (Contributed by Jarvin Udandy, 6-Sep-2020.)
(πœ“ β†’ πœ‘)    &   (πœ“ β†’ Β¬ (πœ“ β†’ (πœ“ ∧ Β¬ πœ“)))    &   ((πœ“ β†’ πœ‘) β†’ ((πœ“ β†’ πœ‘) β†’ πœ‘))    β‡’   (πœ“ β†’ (Β¬ (πœ“ β†’ (πœ“ ∧ Β¬ πœ“)) ↔ (πœ“ β†’ πœ‘)))
 
Theoremconfun3 45651 Confun's more complex form where both a,d have been "defined". (Contributed by Jarvin Udandy, 6-Sep-2020.)
(πœ‘ ↔ (πœ’ β†’ πœ“))    &   (πœƒ ↔ Β¬ (πœ’ β†’ (πœ’ ∧ Β¬ πœ’)))    &   (πœ’ β†’ πœ“)    &   (πœ’ β†’ Β¬ (πœ’ β†’ (πœ’ ∧ Β¬ πœ’)))    &   ((πœ’ β†’ πœ“) β†’ ((πœ’ β†’ πœ“) β†’ πœ“))    β‡’   (πœ’ β†’ (Β¬ (πœ’ β†’ (πœ’ ∧ Β¬ πœ’)) ↔ (πœ’ β†’ πœ“)))
 
Theoremconfun4 45652 An attempt at derivative. Resisted simplest path to a proof. (Contributed by Jarvin Udandy, 6-Sep-2020.)
πœ‘    &   ((πœ‘ β†’ πœ“) β†’ πœ“)    &   (πœ“ β†’ (πœ‘ β†’ πœ’))    &   ((πœ’ β†’ πœƒ) β†’ ((πœ‘ β†’ πœƒ) ↔ πœ“))    &   (𝜏 ↔ (πœ’ β†’ πœƒ))    &   (πœ‚ ↔ Β¬ (πœ’ β†’ (πœ’ ∧ Β¬ πœ’)))    &   πœ“    &   (πœ’ β†’ πœƒ)    β‡’   (πœ’ β†’ (πœ“ β†’ 𝜏))
 
Theoremconfun5 45653 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.)
πœ‘    &   ((πœ‘ β†’ πœ“) β†’ πœ“)    &   (πœ“ β†’ (πœ‘ β†’ πœ’))    &   ((πœ’ β†’ πœƒ) β†’ ((πœ‘ β†’ πœƒ) ↔ πœ“))    &   (𝜏 ↔ (πœ’ β†’ πœƒ))    &   (πœ‚ ↔ Β¬ (πœ’ β†’ (πœ’ ∧ Β¬ πœ’)))    &   πœ“    &   (πœ’ β†’ πœƒ)    β‡’   (πœ’ β†’ (πœ‚ ↔ 𝜏))
 
Theoremplcofph 45654 Given, a,b and a "definition" for c, c is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(πœ’ ↔ ((((πœ‘ ∧ πœ“) ↔ πœ‘) β†’ (πœ‘ ∧ Β¬ (πœ‘ ∧ Β¬ πœ‘))) ∧ (πœ‘ ∧ Β¬ (πœ‘ ∧ Β¬ πœ‘))))    &   πœ‘    &   πœ“    β‡’   πœ’
 
Theorempldofph 45655 Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(𝜏 ↔ ((πœ’ β†’ πœƒ) ∧ (πœ‘ ↔ πœ’) ∧ ((πœ‘ β†’ πœ“) β†’ (πœ“ ↔ πœƒ))))    &   πœ‘    &   πœ“    &   πœ’    &   πœƒ    β‡’   πœ
 
Theoremplvcofph 45656 Given, a,b,d, and "definitions" for c, e, f: f is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(πœ’ ↔ ((((πœ‘ ∧ πœ“) ↔ πœ‘) β†’ (πœ‘ ∧ Β¬ (πœ‘ ∧ Β¬ πœ‘))) ∧ (πœ‘ ∧ Β¬ (πœ‘ ∧ Β¬ πœ‘))))    &   (𝜏 ↔ ((πœ’ β†’ πœƒ) ∧ (πœ‘ ↔ πœ’) ∧ ((πœ‘ β†’ πœ“) β†’ (πœ“ ↔ πœƒ))))    &   (πœ‚ ↔ (πœ’ ∧ 𝜏))    &   πœ‘    &   πœ“    &   πœƒ    β‡’   πœ‚
 
Theoremplvcofphax 45657 Given, a,b,d, and "definitions" for c, e, f, g: g is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
(πœ’ ↔ ((((πœ‘ ∧ πœ“) ↔ πœ‘) β†’ (πœ‘ ∧ Β¬ (πœ‘ ∧ Β¬ πœ‘))) ∧ (πœ‘ ∧ Β¬ (πœ‘ ∧ Β¬ πœ‘))))    &   (𝜏 ↔ ((πœ’ β†’ πœƒ) ∧ (πœ‘ ↔ πœ’) ∧ ((πœ‘ β†’ πœ“) β†’ (πœ“ ↔ πœƒ))))    &   (πœ‚ ↔ (πœ’ ∧ 𝜏))    &   πœ‘    &   πœ“    &   πœƒ    &   (𝜁 ↔ Β¬ (πœ“ ∧ Β¬ 𝜏))    β‡’   πœ
 
Theoremplvofpos 45658 rh is derivable because ONLY one of ch, th, ta, et is implied by mu. (Contributed by Jarvin Udandy, 11-Sep-2020.)
(πœ’ ↔ (Β¬ πœ‘ ∧ Β¬ πœ“))    &   (πœƒ ↔ (Β¬ πœ‘ ∧ πœ“))    &   (𝜏 ↔ (πœ‘ ∧ Β¬ πœ“))    &   (πœ‚ ↔ (πœ‘ ∧ πœ“))    &   (𝜁 ↔ (((((Β¬ ((πœ‡ β†’ πœ’) ∧ (πœ‡ β†’ πœƒ)) ∧ Β¬ ((πœ‡ β†’ πœ’) ∧ (πœ‡ β†’ 𝜏))) ∧ Β¬ ((πœ‡ β†’ πœ’) ∧ (πœ’ β†’ πœ‚))) ∧ Β¬ ((πœ‡ β†’ πœƒ) ∧ (πœ‡ β†’ 𝜏))) ∧ Β¬ ((πœ‡ β†’ πœƒ) ∧ (πœ‡ β†’ πœ‚))) ∧ Β¬ ((πœ‡ β†’ 𝜏) ∧ (πœ‡ β†’ πœ‚))))    &   (𝜎 ↔ (((πœ‡ β†’ πœ’) ∨ (πœ‡ β†’ πœƒ)) ∨ ((πœ‡ β†’ 𝜏) ∨ (πœ‡ β†’ πœ‚))))    &   (𝜌 ↔ (𝜁 ∧ 𝜎))    &   πœ    &   πœŽ    β‡’   πœŒ
 
Theoremmdandyv0 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv1 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv2 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv3 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv4 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv5 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv6 45665 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv7 45666 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ βŠ₯)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ‘))
 
Theoremmdandyv8 45667 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv9 45668 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv10 45669 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv11 45670 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ βŠ₯)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ‘)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv12 45671 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv13 45672 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ βŠ₯)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ‘)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv14 45673 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ βŠ₯)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ‘) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyv15 45674 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.)
(πœ‘ ↔ βŠ₯)    &   (πœ“ ↔ ⊀)    &   (πœ’ ↔ ⊀)    &   (πœƒ ↔ ⊀)    &   (𝜏 ↔ ⊀)    &   (πœ‚ ↔ ⊀)    β‡’   ((((πœ’ ↔ πœ“) ∧ (πœƒ ↔ πœ“)) ∧ (𝜏 ↔ πœ“)) ∧ (πœ‚ ↔ πœ“))
 
Theoremmdandyvr0 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr1 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr2 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr3 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr4 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr5 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr6 45681 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr7 45682 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜁))
 
Theoremmdandyvr8 45683 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr9 45684 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr10 45685 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr11 45686 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜁)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr12 45687 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr13 45688 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜁)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr14 45689 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜁) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvr15 45690 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.)
(πœ‘ ↔ 𝜁)    &   (πœ“ ↔ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ↔ 𝜎) ∧ (πœƒ ↔ 𝜎)) ∧ (𝜏 ↔ 𝜎)) ∧ (πœ‚ ↔ 𝜎))
 
Theoremmdandyvrx0 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜁) ∧ (πœƒ ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx1 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜎) ∧ (πœƒ ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx2 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜁) ∧ (πœƒ ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜁)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx3 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜎) ∧ (πœƒ ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜁)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx4 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜁) ∧ (πœƒ ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜎)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx5 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜎) ∧ (πœƒ ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜎)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx6 45697 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜁) ∧ (πœƒ ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜎)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx7 45698 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ“)    &   (𝜏 ↔ πœ“)    &   (πœ‚ ↔ πœ‘)    β‡’   ((((πœ’ ⊻ 𝜎) ∧ (πœƒ ⊻ 𝜎)) ∧ (𝜏 ⊻ 𝜎)) ∧ (πœ‚ ⊻ 𝜁))
 
Theoremmdandyvrx8 45699 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ‘)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ⊻ 𝜁) ∧ (πœƒ ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (πœ‚ ⊻ 𝜎))
 
Theoremmdandyvrx9 45700 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.)
(πœ‘ ⊻ 𝜁)    &   (πœ“ ⊻ 𝜎)    &   (πœ’ ↔ πœ“)    &   (πœƒ ↔ πœ‘)    &   (𝜏 ↔ πœ‘)    &   (πœ‚ ↔ πœ“)    β‡’   ((((πœ’ ⊻ 𝜎) ∧ (πœƒ ⊻ 𝜁)) ∧ (𝜏 ⊻ 𝜁)) ∧ (πœ‚ ⊻ 𝜎))
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