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Proof step 88 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(m, bits2) = [], bits3 = [], bits2 = [], fillsgn(m, bitsand(fillsgn(m, bits2), bits3)) = fillsgn(m, bitsand(bits2, bits3))

User applied rule apply induction

And closed the goal!

Rule argument

subst: [n, bits, bits1] → [m, bits2, bits3]

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: m, bits2, bits3