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Proof step 68 from proof and-sgn (previous step)

(¬ boolvar0) ∈ bits0, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = [], sgn(bitsand(sgn(bits0) + boolvar0 ', bits2 + boolvar ')) = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits0 + boolvar0 ' = bits0.first '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 69:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 70:
1 < #(sgn(bits0) + boolvar0 ') ∧ 1 < #(bits2 + boolvar '), (¬ boolvar0) ∈ bits0, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = [], sgn(bitsand((sgn(bits0) + boolvar0 ').butlast, (bits2 + boolvar ').butlast) + ((sgn(bits0) + boolvar0 ').last ∧ (bits2 + boolvar ').last)) = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits0 + boolvar0 ' = bits0.first '

Rule argument

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

Additional information

Used simplifier rules: ⊦ #(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 = [];% ⊦ sgn(bits) = [] ↔ bits = [];% ⊦ boolvar ' = bits + boolvar0 ' ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ [] + bits = bits;% ⊦ boolvar '.first ↔ boolvar; Used terms: bits2 + boolvar ', sgn(bits0) + boolvar0 '