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Proof step 3 from proof bitsand-fillsgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ 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 = []

Proof step 3 from proof bitsand-fillsgn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information