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Proof step 33 from proof bib-on-bits (previous step)

# int2bits(i) < 8, i ≤ 0, ~1 ≤ i ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i))))) = bits2int(lastn(8, fillsgn(8, int2bits(i)))), i ∈byte

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = 0

Additional information

Used simplifier rules: inbyte: ⊦ i ∈byte ↔ ~128 ≤ i ∧ i ≤ 127;%inbyte: ⊦ i ∈byte ↔ ~128 ≤ i ∧ i ≤ 127;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ int2bits(0) = false ';% ⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n);%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ mklist(boolvar, 0) = [];% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%0 < n ⊦ lastn((* n), bits + boolvar ') = lastn(n -1, bits) + boolvar ';% ⊦ lastn(1, boolvar ') = boolvar ';% ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ bits2uint(false ') = 0;% ⊦ # boolvar ' = 0 +1; Used formulas: i = 0