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Proof step 10 from proof bits2uint-not-neg (previous step)
| ∀ bits1. # bits1 < #(bits0 + false ') → true → (bits2uint(bits1) ≤ 0 ↔ bits2uint(bits1) = 0) ⊦ bits2uint(bits0) ≤ 0 ↔ bits2uint(bits0) = 0 |
User applied rule apply induction
And closed the goal!
Rule argument
subst: [bits1] → [bits0]Additional information
Used simplifier rules:
⊦ ¬ 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