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Proof step 10 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 = [], bits0 = true ', false ∈ bitsand(bits, bits0), # bits ≠ 1 ∧ # bits0 ≠ 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ bits ≠ [] ∧ bits0 = false ' → bitsand(bits, bits0) = false '
Proof step 12:
¬(bits ≠ [] ∧ bits0 = false '), false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < # bits → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits = [], bits0 = [], bits0 = true ', false ∈ bitsand(bits, bits0), # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

name: and-rec-02 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: bits0, bits