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Proof step 69 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ # bits0 = 1, bits0 = [], fillsgn(n, bitsand(fillsgn(n, bits2 + boolvar '), bits0)) = fillsgn(n, bitsand(bits2, bits0.butlast) + (boolvar ∧ bits0.last)'), bits2 = []

User applied rule insert elim lemma

And obtained the following new goals

Proof step 70:
bits0 = bits3 + boolvar0 ', ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ # bits0 = 1, bits0 = [], fillsgn(n, bitsand(fillsgn(n, bits2 + boolvar '), bits0)) = fillsgn(n, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), bits2 = []

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits0, boolvar0, bits3] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: bits0, boolvar0, bits3