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Proof step 9 from proof sis-on-bits (previous step)

¬ # int2bits(i) ≤ 16, i ≤ 32767, ~32768 ≤ i ⊦ 16 = # fillsgn(16, int2bits(i)), i = bits2int(lastn(16, fillsgn(16, int2bits(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
16 < # int2bits(i), ~32768 ≤ i, i ≤ 32767 ⊦ i = bits2int(lastn(16, int2bits(i))), 16 = # fillsgn(16, int2bits(i))

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < # bits → lastn(n, fillsgn(n, bits)) = lastn(n, bits);% ⊦ ¬ m ≤ n ↔ n < m;