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Proof step 41 from proof bitsand-fillsgn (previous step)

0 < n, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(mklist(true, n), bits0)) = fillsgn(n, bits0), bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(mklist(true, n), bits0)) = fillsgn(n, bits0), bits0 = [], n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < n ↔ n ≠ 0;