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Proof step 74 from proof bitsand-fillsgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 75:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)

Rule argument

name: and-rec rotate: F subst: [bits0, bits] → [bits3 + boolvar0 ', bits2 + boolvar ']

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + boolvar ').butlast = bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m + 1 ≠ 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = []; Used terms: bits3 + boolvar0 ', bits2 + boolvar '