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Proof step 25 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)) + (false ' + lastn(i→n(i), uint2bits(j))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# lastn(i→n(i), uint2bits(j)) = i→n(i), i→n(i) < # uint2bits(j), 2 ^ i < j, 0 ≤ i, 2 ^ i ≤ j, i→n(i) + 1 < # uint2bits(j) ⊦ uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + lastn(i→n(i), uint2bits(j))

Rule argument

Additional information

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