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Proof step 13 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 ≠ bits.first ' ∧ bits0 ≠ bits0.first '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsand(bits, bits0) = bits0
Proof step 15:
¬(bits0 ≠ [] ∧ bits = true '), ∀ 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 ≠ bits.first ' ∧ bits0 ≠ bits0.first '

Rule argument

name: and-rec-03 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;% ⊦ sgn(boolvar ') = boolvar '; Used terms: bits0, bits