S = 10 side=2 [a=10, b=10, c>10..]
a=10 {history=S} side=1 [S=10,g=10,…]
g=10 {history=S,a} side=0 [z=8,S=10,…]
z=8 {} side=2
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!P ^ Q -> (!P) ^ Q; !(P ^ Q)
!P<=>Q=>WvS^T -> (!P)<=>[Q=>[Wv(S^T)]]
Color(WA, R) v Color(WA, G) v Color(WA, B)
Color(WA, R) => ![Color(NT, R) v Color(SA, R)]
Color(WA, G) => ![Color(NT, G) v Color(SA, G)]
Color(WA, B) => ![Color(NT, B) v Color(SA, B)]
Color(WA, R) => ![Color(WA, B) v Color(WA, G)]
Color(WA, G) => ![Color(WA, R) v Color(WA, B)]
Color(WA, B) => ![Color(WA, R) v Color(WA, G)]
Color(NSW, R) => ![Color(Q, R) v Color(SA, R) v Color(V, R)]
S = {Color(WA, R)=true; Color(WA, G) = true; …} A v (b ^ c) -> (a v b) ^ (a v c)
Delete: P v !P v Q v W
C ^ A => (B v D) C^D
Step 2 =>:
!(C ^ A) v (B v D) C^D
(!C v !A) v (B v D) C^D
Step 5 split:
!C !A B D C
([A | (B & C)] => (P & ! Q)) => W
[(x+y)+z]+w
P & ! Q => W <=> A | B & C
((P & ! Q => W) => A | B & C) & (A | B & C => (P & ! Q => W))
(!(![P & ! Q] v W) v [A | B & C]) & (![A | B & C] v (![P & ! Q] v W))
(([P & !Q] & !W) v [A v B & C]) & ([!A & (!B v !C)] v ([!P v Q] v W))
1. [([P & !Q] & !W) v [A v (B & C)] -> note was done differently [!A & (!B v !C)] v ([!P v Q] v W)
1. [([P & !Q] & !W) v A] v B [([P & !Q] & !W) v A] v C
[([P & !Q] v A) & (!W v A)] v B [([P & !Q] v A) & (!W v A)] v C
(([P & !Q] v A) v B) !W v A v B
(([P & !Q] v A) v C) !W v A v C
[(P v A) ^ (!Q v A)] v B
!W v A v B !W v A v C PvAvB !Q v A v B PvAvC !Q v A v C
[!A & (!B v !C)] v ([!P v Q] v W)
!A v !P v Q v W
!B v !C v !P v Q v W !W v A v B
!W v A v C PvAvB
!Q v A v B
!Q v A v C
Assign ={A:unbound, B: unbound, C: unbound, D: unbound}
!C !A B D C
!A B D D !B
C=true, A=false
C=true, A=false, B=false, D=false (default)
Harder DPLL B A P !W
C A !Q !W !P W !B !A Q W !B !A !P W !C !A Q W !C !A !AB
!P W !B Q W !B !P W !C Q W !C B
A=true, B=true !P W
A=true, B=true,C=false
A=true, B=true,C=false, W=true, P=false, Q=false
!A v !P v Q v W
!B v !C v !P v Q v W !W v A v B
!W v A v C PvAvB
!Q v A v B
!Q v A v C
Guess: A=false
!B v !C v !P v Q v W !W v B
!Q v B PvC !Q v C
Guess: B=false
!Q PvC !Q v C
Easy: P=true
At this point I changed several signs to demonstrate a contradiction and backtracking.
Q=true {0}
At this point you fail back to the last hard-case (not easy cases). So B=false, start from that state and try B=true, if that fails, you go all the way back to the guess of A=false, and try instead A=true.
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