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Proof step 11 from proof and-size-one (previous step)

1 ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # true ' → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits1 = [], bitsand(true ', bits0 + bits1) = bits0 + bits1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsand(bits, bits0) = bits0

Rule argument

name: and-rec-03 rotate: F subst: [bits0, bits] → [bits0 + bits1, true ']

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits0 + bits1, true '