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Proof step 16 from proof int2bits-leneq-pos (previous step)

0 < j ∧ 0 ≤ n→i(# int2bits(j)) - 1, 0 < j, ¬ # int2bits(j) ≤ i→n(n→i(# int2bits(j)) - 1) + 1 ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n; from specification intnat%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs