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Proof step 4 from proof and-zerosin (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 6:
¬(1 < # bits ∧ 1 < # bits0), true ∈ bitsand(bits, bits0), ∀ bits1, bits0. # bits1 < # bits → true ∈ bitsand(bits1, bits0) → bits1 = [] ∨ bits0 = [] ∨ true ∈ bits1 ⊦ bits0 = [], bits = [], true ∈ bits
Proof step 22:
1 < # bits ∧ 1 < # bits0, true ∈ bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last), ∀ bits1, bits0. # bits1 < # bits → true ∈ bitsand(bits1, bits0) → bits1 = [] ∨ bits0 = [] ∨ true ∈ bits1 ⊦ bits0 = [], bits = [], true ∈ bits

Rule argument

name: and-rec rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used terms: bits0, bits