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Proof step 38 from proof bitsand-fillsgn (previous step)

bits0 ≠ [] ∧ bits = true ', # bits = 1, ∀ 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, bits0), bits0 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
∀ 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, true '), bits0)) = fillsgn(n, bits0), bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ # boolvar ' = 0 +1;