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Proof step 16 from proof and-onesin (previous step)

∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0, bits0 = [], bits = [], bits0 = true ', bits0 = false ', bits = true ', false ∈ bits, bits ≠ bits.first ' ∧ bits0 ≠ bits0.first '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ bits0 ≠ [] ∧ bits = false ' → bitsand(bits, bits0) = false '
Proof step 18:
¬(bits0 ≠ [] ∧ bits = false '), ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0, bits0 = [], bits = [], bits0 = true ', bits0 = false ', bits = true ', false ∈ bits, bits ≠ bits.first ' ∧ bits0 ≠ bits0.first '

Rule argument

name: and-rec-04 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

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