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Theorem mdandysum2p2e4 47888
Description: 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.)

Hypotheses
Ref Expression
mdandysum2p2e4.1 (jth ↔ ⊥)
mdandysum2p2e4.2 (jta ↔ ⊤)
mdandysum2p2e4.a (𝜑 ↔ (𝜃𝜏))
mdandysum2p2e4.b (𝜓 ↔ (𝜂𝜁))
mdandysum2p2e4.c (𝜒 ↔ (𝜎𝜌))
mdandysum2p2e4.d (𝜃jth)
mdandysum2p2e4.e (𝜏jth)
mdandysum2p2e4.f (𝜂jta)
mdandysum2p2e4.g (𝜁jta)
mdandysum2p2e4.h (𝜎jth)
mdandysum2p2e4.i (𝜌jth)
mdandysum2p2e4.j (𝜇jth)
mdandysum2p2e4.k (𝜆jth)
mdandysum2p2e4.l (𝜅 ↔ ((𝜃𝜏) ⊻ (𝜃𝜏)))
mdandysum2p2e4.m (jph ↔ ((𝜂𝜁) ∨ 𝜑))
mdandysum2p2e4.n (jps ↔ ((𝜎𝜌) ∨ 𝜓))
mdandysum2p2e4.o (jch ↔ ((𝜇𝜆) ∨ 𝜒))
Assertion
Ref Expression
mdandysum2p2e4 ((((((((((((((((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 ↔ ⊥)))

Proof of Theorem mdandysum2p2e4
StepHypRef Expression
1 mdandysum2p2e4.a . 2 (𝜑 ↔ (𝜃𝜏))
2 mdandysum2p2e4.b . 2 (𝜓 ↔ (𝜂𝜁))
3 mdandysum2p2e4.c . 2 (𝜒 ↔ (𝜎𝜌))
4 mdandysum2p2e4.d . . 3 (𝜃jth)
5 mdandysum2p2e4.1 . . 3 (jth ↔ ⊥)
64, 5aisbbisfaisf 47791 . 2 (𝜃 ↔ ⊥)
7 mdandysum2p2e4.e . . 3 (𝜏jth)
87, 5aisbbisfaisf 47791 . 2 (𝜏 ↔ ⊥)
9 mdandysum2p2e4.f . . 3 (𝜂jta)
10 mdandysum2p2e4.2 . . 3 (jta ↔ ⊤)
119, 10aiffbbtat 47790 . 2 (𝜂 ↔ ⊤)
12 mdandysum2p2e4.g . . 3 (𝜁jta)
1312, 10aiffbbtat 47790 . 2 (𝜁 ↔ ⊤)
14 mdandysum2p2e4.h . . 3 (𝜎jth)
1514, 5aisbbisfaisf 47791 . 2 (𝜎 ↔ ⊥)
16 mdandysum2p2e4.i . . 3 (𝜌jth)
1716, 5aisbbisfaisf 47791 . 2 (𝜌 ↔ ⊥)
18 mdandysum2p2e4.j . . 3 (𝜇jth)
1918, 5aisbbisfaisf 47791 . 2 (𝜇 ↔ ⊥)
20 mdandysum2p2e4.k . . 3 (𝜆jth)
2120, 5aisbbisfaisf 47791 . 2 (𝜆 ↔ ⊥)
22 mdandysum2p2e4.l . 2 (𝜅 ↔ ((𝜃𝜏) ⊻ (𝜃𝜏)))
23 mdandysum2p2e4.m . 2 (jph ↔ ((𝜂𝜁) ∨ 𝜑))
24 mdandysum2p2e4.n . 2 (jps ↔ ((𝜎𝜌) ∨ 𝜓))
25 mdandysum2p2e4.o . 2 (jch ↔ ((𝜇𝜆) ∨ 𝜒))
261, 2, 3, 6, 8, 11, 13, 15, 17, 19, 21, 22, 23, 24, 25dandysum2p2e4 47887 1 ((((((((((((((((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 ↔ ⊥)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wb 209  wa 401  wo 861  wxo 1541  wtru 1571  wfal 1582
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-xor 1542  df-tru 1573  df-fal 1583
This theorem is used by: (None)
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