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Proof step 3 from proof int2bits-lslen-12 (previous step)

2 ^ j < i, 0 < i, 0 ≤ j ⊦ ¬ # int2bits(i) < i→n(j) + 2 ∧ i ≠ 2 ^ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# int2bits(i) < i→n(j) + 2, 0 ≤ j, 0 < i, 2 ^ j < i ⊦ i = 2 ^ j

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ i < i;% ⊦ 0 ≤ j → (2 ^ j < 2 ^ j0 ↔ j < j0);% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;