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Proof step 67 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))) ⊦ bitsand(bitsinvert(bits0), bits2) = [], bitsand(bitsinvert(bits2), bits0) = [], bits0 = [], bits2 = [], bits2int(bitsxor(bits0, bits2) + (¬(boolvar0 ↔ boolvar))') = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)')

User applied rule apply induction

And obtained the following new goals

Proof step 68:
bits0 = [] ∨ bits2 = [] ∨ bits2int(bitsxor(bits0, bits2)) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0))), ∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bitsand(bitsinvert(bits0), bits2) = [], bitsand(bitsinvert(bits2), bits0) = [], bits0 = [], bits2 = [], bits2int(bitsxor(bits0, bits2) + (¬(boolvar0 ↔ boolvar))') = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)')

Rule argument

subst: [bits, bits1] → [bits0, bits2]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits0, bits2