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Proof step 64 from proof bitsxor-is-orand (previous step)

∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits0 = [], bits2 = [], bits2int(bitsxor(bits0 + boolvar0 ', bits2 + boolvar ')) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)'), bitsand(bitsinvert(bits0), bits2) = [], bitsand(bitsinvert(bits2), bits0) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: bitsxor-rec-01 rotate: F subst: [bits0, bits] → [bits2 + boolvar ', 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 = []; Used terms: bits2 + boolvar ', bits0 + boolvar0 '