Statement List for Quantum Logic Explorer - 501-600 - Page 6 of 11
| Type | Label | Description |
| Statement |
| |
| Theorem | binr3 501 |
Pavicic binary logic axr3 analog.
|
| (a
→3 c) =
1 & (b →3 c) = 1
⇒ ((a ∪
b) →3 c) = 1 |
| |
| Theorem | i31 502 |
Theorem for Kalmbach implication.
|
| (a
→3 1) = 1 |
| |
| Theorem | i3aa 503 |
Add antecedent.
|
| a
= 1 ⇒ (b →3 a) = 1 |
| |
| Theorem | i3abs1 504 |
Antecedent absorption.
|
| (a
→3 (a →3
(a →3 b))) = (a
→3 (a →3
b)) |
| |
| Theorem | i3abs2 505 |
Antecedent absorption.
|
| (a
→3 (a →3
(a →3 b))) = 1
⇒ (a →3
(a →3 b)) = 1 |
| |
| Theorem | i3abs3 506 |
Antecedent absorption.
|
| ((a →3 b) →3 ((a →3 b) →3 a)) = ((a
→3 b) →3
a) |
| |
| Theorem | i3orcom 507 |
Commutative law for conjunction with Kalmbach implication.
|
| ((a ∪ b)
→3 (b ∪ a)) = 1 |
| |
| Theorem | i3ancom 508 |
Commutative law for disjunction with Kalmbach implication.
|
| ((a ∩ b)
→3 (b ∩ a)) = 1 |
| |
| Theorem | bi3tr 509 |
Transitive inference.
|
| a
= b
& (b →3
c) = 1
⇒ (a →3
c) = 1 |
| |
| Theorem | i3btr 510 |
Transitive inference.
|
| (a
→3 b) =
1 & b = c
⇒ (a →3
c) = 1 |
| |
| Theorem | i33tr1 511 |
Transitive inference useful for introducing definitions.
|
| (a
→3 b) =
1 & c = a & d = b
⇒ (c →3
d) = 1 |
| |
| Theorem | i33tr2 512 |
Transitive inference useful for eliminating definitions.
|
| (a
→3 b) =
1 & a = c & b = d
⇒ (c →3
d) = 1 |
| |
| Theorem | i3con1 513 |
Contrapositive.
|
| (a⊥ →3 b⊥ ) = 1
⇒ (b →3
a) = 1 |
| |
| Theorem | i3ror 514 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 ⇒ ((a ∪ c)
→3 (b ∪ c)) = 1 |
| |
| Theorem | i3lor 515 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 ⇒ ((c ∪ a)
→3 (c ∪ b)) = 1 |
| |
| Theorem | i32or 516 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 & (c →3 d) = 1
⇒ ((a ∪
c) →3 (b ∪ d)) =
1 |
| |
| Theorem | i3ran 517 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 ⇒ ((a ∩ c)
→3 (b ∩ c)) = 1 |
| |
| Theorem | i3lan 518 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 ⇒ ((c ∩ a)
→3 (c ∩ b)) = 1 |
| |
| Theorem | i32an 519 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 & (c →3 d) = 1
⇒ ((a ∩
c) →3 (b ∩ d)) =
1 |
| |
| Theorem | i3ri3 520 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 & (b →3 a) = 1
⇒ ((a →3
c) →3 (b →3 c)) = 1 |
| |
| Theorem | i3li3 521 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 & (b →3 a) = 1
⇒ ((c →3
a) →3 (c →3 b)) = 1 |
| |
| Theorem | i32i3 522 |
WQL (Weak Quantum Logic) rule.
|
| (a
→3 b) =
1 & (b →3 a) = 1
& (c →3
d) = 1
& (d →3
c) = 1
⇒ ((a →3
c) →3 (b →3 d)) = 1 |
| |
| Theorem | i0i3tr 523 |
Transitive inference.
|
| (a
→3 (a →3
b)) = 1
& (b →3
c) = 1
⇒ (a →3
(a →3 c)) = 1 |
| |
| Theorem | i3i0tr 524 |
Transitive inference.
|
| (a
→3 b) =
1 & (b →3 (b →3 c)) = 1
⇒ (a →3
(a →3 c)) = 1 |
| |
| Theorem | i3th1 525 |
Theorem for Kalmbach implication.
|
| (a
→3 (a →3
(b →3 a))) = 1 |
| |
| Theorem | i3th2 526 |
Theorem for Kalmbach implication.
|
| (a
→3 (b →3
(b →3 a))) = 1 |
| |
| Theorem | i3th3 527 |
Theorem for Kalmbach implication.
|
| (a⊥ →3 (a →3 (a →3 b))) = 1 |
| |
| Theorem | i3th4 528 |
Theorem for Kalmbach implication.
|
| (a
→3 (b →3
b)) = 1 |
| |
| Theorem | i3th5 529 |
Theorem for Kalmbach implication.
|
| ((a →3 b) →3 (a →3 (a →3 b))) = 1 |
| |
| Theorem | i3th6 530 |
Theorem for Kalmbach implication.
|
| ((a →3 (a →3 (a →3 b))) →3 (a →3 (a →3 b))) = 1 |
| |
| Theorem | i3th7 531 |
Theorem for Kalmbach implication.
|
| (a
→3 ((a →3
b) →3 a)) = 1 |
| |
| Theorem | i3th8 532 |
Theorem for Kalmbach implication.
|
| ((a →3 b)⊥ →3 ((a →3 b) →3 a)) = 1 |
| |
| Theorem | i3con 533 |
Theorem for Kalmbach implication.
|
| ((a →3 b) →3 ((a →3 b) →3 (b⊥ →3 a⊥ ))) = 1 |
| |
| Theorem | i3orlem1 534 |
Lemma for Kalmbach implication OR builder.
|
| ((a ∪ c)
∩ ((a ∪ c)⊥ ∪ (b ∪ c)))
≤ ((a ∪ c) →3 (b ∪ c)) |
| |
| Theorem | i3orlem2 535 |
Lemma for Kalmbach implication OR builder.
|
| (a
∩ b) ≤ ((a ∪ c)
→3 (b ∪ c)) |
| |
| Theorem | i3orlem3 536 |
Lemma for Kalmbach implication OR builder.
|
| c
≤ ((a ∪ c) →3 (b ∪ c)) |
| |
| Theorem | i3orlem4 537 |
Lemma for Kalmbach implication OR builder.
|
| ((a ∪ c)⊥ ∩ (b ∪ c))
≤ ((a ∪ c) →3 (b ∪ c)) |
| |
| Theorem | i3orlem5 538 |
Lemma for Kalmbach implication OR builder.
|
| ((a⊥ ∩ b⊥ ) ∩ c⊥ ) ≤ ((a ∪ c)
→3 (b ∪ c)) |
| |
| Theorem | i3orlem6 539 |
Lemma for Kalmbach implication OR builder.
|
| ((a →3 b)⊥ ∪ ((a ∪ c)
→3 (b ∪ c))) = (((a
∪ b) ∩ (a⊥ →3 b⊥ )) ∪ ((a ∪ c)
→3 (b ∪ c))) |
| |
| Theorem | i3orlem7 540 |
Lemma for Kalmbach implication OR builder.
|
| (a
∩ b⊥ ) ≤
((a →3 b)⊥ ∪ ((a ∪ c)
→3 (b ∪ c))) |
| |
| Theorem | i3orlem8 541 |
Lemma for Kalmbach implication OR builder.
|
| (((a ∪ b)
∩ (a ∪ b⊥ )) ∩ a⊥ ) ≤ ((a →3 b)⊥ ∪ ((a ∪ c)
→3 (b ∪ c))) |
| |
| Theorem | ud1lem1 542 |
Lemma for unified disjunction.
|
| ((a →1 b) →1 (b →1 a)) = (a ∪
(a⊥ ∩ b⊥ )) |
| |
| Theorem | ud1lem2 543 |
Lemma for unified disjunction.
|
| ((a ∪ (a⊥ ∩ b⊥ )) →1 a) = (a ∪
b) |
| |
| Theorem | ud1lem3 544 |
Lemma for unified disjunction.
|
| ((a →1 b) →1 (a ∪ b)) =
(a ∪ b) |
| |
| Theorem | ud2lem1 545 |
Lemma for unified disjunction.
|
| ((a →2 b) →2 (b →2 a)) = (a ∪
(a⊥ ∩ b⊥ )) |
| |
| Theorem | ud2lem2 546 |
Lemma for unified disjunction.
|
| ((a ∪ (a⊥ ∩ b⊥ )) →2 a) = (a ∪
b) |
| |
| Theorem | ud2lem3 547 |
Lemma for unified disjunction.
|
| ((a →2 b) →2 (a ∪ b)) =
(a ∪ b) |
| |
| Theorem | ud3lem1a 548 |
Lemma for unified disjunction.
|
| ((a →3 b)⊥ ∩ (b →3 a)) = (a ∩
b⊥ ) |
| |
| Theorem | ud3lem1b 549 |
Lemma for unified disjunction.
|
| ((a →3 b)⊥ ∩ (b →3 a)⊥ ) = 0 |
| |
| Theorem | ud3lem1c 550 |
Lemma for unified disjunction.
|
| ((a →3 b)⊥ ∪ (b →3 a)) = (a ∪
b⊥ ) |
| |
| Theorem | ud3lem1d 551 |
Lemma for unified disjunction.
|
| ((a →3 b) ∩ ((a
→3 b)⊥
∪ (b →3 a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))) |
| |
| Theorem | ud3lem1 552 |
Lemma for unified disjunction.
|
| ((a →3 b) →3 (b →3 a)) = (a ∪
(a⊥ ∩ b⊥ )) |
| |
| Theorem | ud3lem2 553 |
Lemma for unified disjunction.
|
| ((a ∪ (a⊥ ∩ b⊥ )) →3 a) = (a ∪
b) |
| |
| Theorem | ud3lem3a 554 |
Lemma for unified disjunction.
|
| ((a →3 b)⊥ ∩ (a ∪ b)) =
(a →3 b)⊥ |
| |
| Theorem | ud3lem3b 555 |
Lemma for unified disjunction.
|
| ((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = 0 |
| |
| Theorem | ud3lem3c 556 |
Lemma for unified disjunction.
|
| ((a →3 b)⊥ ∪ (a ∪ b)) =
(a ∪ b) |
| |
| Theorem | ud3lem3d 557 |
Lemma for unified disjunction.
|
| ((a →3 b) ∩ ((a
→3 b)⊥
∪ (a ∪ b))) = ((a⊥ ∩ b) ∪ (a
∩ (a⊥ ∪ b))) |
| |
| Theorem | ud3lem3 558 |
Lemma for unified disjunction.
|
| ((a →3 b) →3 (a ∪ b)) =
(a ∪ b) |
| |
| Theorem | ud4lem1a 559 |
Lemma for unified disjunction.
|
| ((a →4 b) ∩ (b
→4 a)) = ((a ∩ b)
∪ (a⊥ ∩ b⊥ )) |
| |
| Theorem | ud4lem1b 560 |
Lemma for unified disjunction.
|
| ((a →4 b)⊥ ∩ (b →4 a)) = (a ∩
b⊥ ) |
| |
| Theorem | ud4lem1c 561 |
Lemma for unified disjunction.
|
| ((a →4 b)⊥ ∪ (b →4 a)) = (a ∪
b⊥ ) |
| |
| Theorem | ud4lem1d 562 |
Lemma for unified disjunction.
|
| (((a →4 b)⊥ ∪ (b →4 a)) ∩ (b
→4 a)⊥ )
= (((a⊥ ∪ b⊥ ) ∩ (a⊥ ∪ b)) ∩ a) |
| |
| Theorem | ud4lem1 563 |
Lemma for unified disjunction.
|
| ((a →4 b) →4 (b →4 a)) = (a ∪
(a⊥ ∩ b⊥ )) |
| |
| Theorem | ud4lem2 564 |
Lemma for unified disjunction.
|
| ((a ∪ (a⊥ ∩ b⊥ )) →4 a) = (a ∪
b) |
| |
| Theorem | ud4lem3a 565 |
Lemma for unified disjunction.
|
| ((a →4 b)⊥ ∩ (a ∪ b)) =
(a →4 b)⊥ |
| |
| Theorem | ud4lem3b 566 |
Lemma for unified disjunction.
|
| ((a →4 b)⊥ ∪ (a ∪ b)) =
(a ∪ b) |
| |
| Theorem | ud4lem3 567 |
Lemma for unified disjunction.
|
| ((a →4 b) →4 (a ∪ b)) =
(a ∪ b) |
| |
| Theorem | ud5lem1a 568 |
Lemma for unified disjunction.
|
| ((a →5 b) ∩ (b
→5 a)) = ((a ∩ b)
∪ (a⊥ ∩ b⊥ )) |
| |
| Theorem | ud5lem1b 569 |
Lemma for unified disjunction.
|
| ((a →5 b)⊥ ∩ (b →5 a)) = (a ∩
b⊥ ) |
| |
| Theorem | ud5lem1c 570 |
Lemma for unified disjunction.
|
| ((a →5 b)⊥ ∩ (b →5 a)⊥ ) = (((a ∪ b)
∩ (a ∪ b⊥ )) ∩ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ ))) |
| |
| Theorem | ud5lem1 571 |
Lemma for unified disjunction.
|
| ((a →5 b) →5 (b →5 a)) = (a ∪
b⊥ ) |
| |
| Theorem | ud5lem2 572 |
Lemma for unified disjunction.
|
| ((a ∪ b⊥ ) →5 a) = (a ∪
(a⊥ ∩ b)) |
| |
| Theorem | ud5lem3a 573 |
Lemma for unified disjunction.
|
| ((a →5 b) ∩ (a
∪ (a⊥ ∩ b))) = ((a
∩ b) ∪ (a⊥ ∩ b)) |
| |
| Theorem | ud5lem3b 574 |
Lemma for unified disjunction.
|
| ((a →5 b)⊥ ∩ (a ∪ (a⊥ ∩ b))) = (a ∩
(a⊥ ∪ b⊥ )) |
| |
| Theorem | ud5lem3c 575 |
Lemma for unified disjunction.
|
| ((a →5 b)⊥ ∩ (a ∪ (a⊥ ∩ b))⊥ ) = (((a ∪ b)
∩ (a ∪ b⊥ )) ∩ a⊥ ) |
| |
| Theorem | ud5lem3 576 |
Lemma for unified disjunction.
|
| ((a →5 b) →5 (a ∪ (a⊥ ∩ b))) = (a ∪
b) |
| |
| Theorem | ud1 577 |
Unified disjunction for Sasaki implication.
|
| (a
∪ b) = ((a →1 b) →1 (((a →1 b) →1 (b →1 a)) →1 a)) |
| |
| Theorem | ud2 578 |
Unified disjunction for Dishkant implication.
|
| (a
∪ b) = ((a →2 b) →2 (((a →2 b) →2 (b →2 a)) →2 a)) |
| |
| Theorem | ud3 579 |
Unified disjunction for Kalmbach implication.
|
| (a
∪ b) = ((a →3 b) →3 (((a →3 b) →3 (b →3 a)) →3 a)) |
| |
| Theorem | ud4 580 |
Unified disjunction for non-tollens implication.
|
| (a
∪ b) = ((a →4 b) →4 (((a →4 b) →4 (b →4 a)) →4 a)) |
| |
| Theorem | ud5 581 |
Unified disjunction for relevance implication.
|
| (a
∪ b) = ((a →5 b) →5 (((a →5 b) →5 (b →5 a)) →5 a)) |
| |
| Theorem | u1lemaa 582 |
Lemma for Sasaki implication study.
|
| ((a →1 b) ∩ a) =
(a ∩ b) |
| |
| Theorem | u2lemaa 583 |
Lemma for Dishkant implication study.
|
| ((a →2 b) ∩ a) =
(a ∩ b) |
| |
| Theorem | u3lemaa 584 |
Lemma for Kalmbach implication study.
|
| ((a →3 b) ∩ a) =
(a ∩ (a⊥ ∪ b)) |
| |
| Theorem | u4lemaa 585 |
Lemma for non-tollens implication study.
|
| ((a →4 b) ∩ a) =
(a ∩ b) |
| |
| Theorem | u5lemaa 586 |
Lemma for relevance implication study.
|
| ((a →5 b) ∩ a) =
(a ∩ b) |
| |
| Theorem | u1lemana 587 |
Lemma for Sasaki implication study.
|
| ((a →1 b) ∩ a⊥ ) = a⊥ |
| |
| Theorem | u2lemana 588 |
Lemma for Dishkant implication study.
|
| ((a →2 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
| |
| Theorem | u3lemana 589 |
Lemma for Kalmbach implication study.
|
| ((a →3 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
| |
| Theorem | u4lemana 590 |
Lemma for non-tollens implication study.
|
| ((a →4 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
| |
| Theorem | u5lemana 591 |
Lemma for relevance implication study.
|
| ((a →5 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
| |
| Theorem | u1lemab 592 |
Lemma for Sasaki implication study.
|
| ((a →1 b) ∩ b) =
((a ∩ b) ∪ (a⊥ ∩ b)) |
| |
| Theorem | u2lemab 593 |
Lemma for Dishkant implication study.
|
| ((a →2 b) ∩ b) =
b |
| |
| Theorem | u3lemab 594 |
Lemma for Kalmbach implication study.
|
| ((a →3 b) ∩ b) =
((a ∩ b) ∪ (a⊥ ∩ b)) |
| |
| Theorem | u4lemab 595 |
Lemma for non-tollens implication study.
|
| ((a →4 b) ∩ b) =
((a ∩ b) ∪ (a⊥ ∩ b)) |
| |
| Theorem | u5lemab 596 |
Lemma for relevance implication study.
|
| ((a →5 b) ∩ b) =
((a ∩ b) ∪ (a⊥ ∩ b)) |
| |
| Theorem | u1lemanb 597 |
Lemma for Sasaki implication study.
|
| ((a →1 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |
| |
| Theorem | u2lemanb 598 |
Lemma for Dishkant implication study.
|
| ((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |
| |
| Theorem | u3lemanb 599 |
Lemma for Kalmbach implication study.
|
| ((a →3 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |
| |
| Theorem | u4lemanb 600 |
Lemma for non-tollens implication study.
|
| ((a →4 b) ∩ b⊥ ) = ((a⊥ ∪ b) ∩ b⊥ ) |