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Proof step 12 from proof bits2uint-not-neg (previous step)

bits2uint(bits0) < 0, ∀ bits1. # bits1 < #(bits0 + true ') → true → (bits2uint(bits1) ≤ 0 ↔ bits2uint(bits1) = 0) ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1] → [bits0]

Additional information

Used simplifier rules: ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ #(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