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Proof step 10 from proof lastn-fill-sgn (previous step)

¬ n ≤ # sgn(bits), n < # bits ⊦ n = # bits, bits = [], lastn(n, fillsgn(n, sgn(bits))) = lastn(n, bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# sgn(bits) < # bits, # sgn(bits) < n, n < # bits, # sgn(bits) + 1 < # bits ⊦ lastn(n, fillsgn(n, sgn(bits))) = lastn(n, bits), bits = [], n = # bits

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1;% ⊦ m < n → n ≠ 0;% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ ¬ m ≤ n ↔ n < m;