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Proof step 24 from proof int2bytes-rec-pos (previous step)

fillsgn(9, uint2bits(i % 256)) ≠ [], uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, 8 < # uint2bits(i / 256) + (# fillsgn(9, uint2bits(i % 256)))-1, 0 ≤ i, 0 < i, 255 < i ⊦ bits2bytes(firstn(# uint2bits(i / 256) + (# fillsgn(9, uint2bits(i % 256)))-1 - 8, uint2bits(i))) + i2b(bits2int(lastn(8, uint2bits(i)))) ' = bits2bytes(uint2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i / j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ n -1 = n - 1;