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Proof step 19 from proof uint2bits-2potrec (previous step)

#(false ' + lastn(i→n(i), uint2bits(j))) ≤ i→n(i) + 1, i→n(i) < # uint2bits(j), i→n(i) + 1 < # uint2bits(j), 2 ^ i ≤ j, 0 ≤ i, 2 ^ i < 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 20:
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

Rule argument

Additional information

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