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Proof step 15 from proof bitsand-fillsgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsand(bits, bits0) = bits

Rule argument

name: and-rec-01 rotate: F subst: [bits0, bits] → [true ', fillsgn(n, bits)]

Additional information

Used simplifier rules: ⊦ fillsgn(n, bits) = [] ↔ bits = [] ∧ n = 0;% ⊦ bits ≠ [] → fillsgn(n, fillsgn(n, bits)) = fillsgn(n, bits); Used terms: true ', fillsgn(n, bits)