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Proof step 4 from proof bitsand-zero-in (previous step)

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

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), false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < # bits → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits = [], bits0 = [], false ∈ bitsand(bits, bits0)
Proof step 15:
1 < # bits ∧ 1 < # bits0, false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < # bits → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits = [], bits0 = [], false ∈ bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)

Rule argument

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

Additional information

Used terms: bits0, bits