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Proof step 6 from proof bitsand-fillsgn (previous step)

¬(1 < # bits ∧ 1 < # bits0), ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = [], # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);