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Proof step 22 from proof uint2bits-2potrec (previous step)

¬ i→n(i) + 1 ≤ #(false ' + lastn(i→n(i), uint2bits(j))), i→n(i) < # uint2bits(j), i→n(i) + 1 < # uint2bits(j), 2 ^ i ≤ j, 0 ≤ i, 2 ^ i < j, # lastn(i→n(i), uint2bits(j)) ≤ i→n(i), # lastn(i→n(i), uint2bits(j)) < # uint2bits(j) ⊦ uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + fillsgn(i→n(i) + 1, false ' + lastn(i→n(i), uint2bits(j))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
2 ^ i ≤ j, i→n(i) + 1 < # uint2bits(j), i→n(i) < # uint2bits(j), # lastn(i→n(i), uint2bits(j)) < # uint2bits(j), # lastn(i→n(i), uint2bits(j)) + 1 < # uint2bits(j), # lastn(i→n(i), uint2bits(j)) < i→n(i), 2 ^ i < j, 0 < i ⊦ # lastn(i→n(i), uint2bits(j)) = i→n(i), uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + fillsgn(i→n(i) + 1, false ' + lastn(i→n(i), uint2bits(j))).rest, i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ m +1 = m + 1;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m ≠ n → (m ≤ n ↔ m < n);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ n → (n ≤ m ↔ m = n);% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;