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

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

Rule argument

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

Additional information

Used simplifier rules: and-zero-02: ⊦ bits ≠ [] → bitsand(false ', bits) = false ';% ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: bits0, bits