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Proof step 52 from proof and-insgn (previous step)

(¬ boolvar1) ∈ bits0, ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(sgn(bits0) + boolvar1 ', bits2 + boolvar0 ') ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first '

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: and-rec rotate: F subst: [bits0, bits] → [bits2 + boolvar0 ', sgn(bits0) + boolvar1 ']

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 = [];% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ [] + bits = bits;% ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: bits2 + boolvar0 ', sgn(bits0) + boolvar1 '