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Proof step 68 from proof bitsand-fillsgn (previous step)

bits = bits2 + boolvar ', ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits0 = [], bits = [], fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits2, bits0.butlast) + (boolvar ∧ bits0.last)'), # bits = 1, # bits0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 69:
∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ # bits0 = 1, bits0 = [], fillsgn(n, bitsand(fillsgn(n, bits2 + boolvar '), bits0)) = fillsgn(n, bitsand(bits2, bits0.butlast) + (boolvar ∧ bits0.last)'), bits2 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];