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Proof step 20 from proof int2bits-leneq-pos (previous step)

2 ^ (n→i(n) - 2) ≤ j ∧ j < 2 ^ (n→i(n) - 1), 0 ≤ j ⊦ # int2bits(j) = n, n = 1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ler: ⊦ m ≤ m; from specification nat%le: ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n; from specification intnat%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%ls-pos-2pot-01: 0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0; from specification int-pot%num-n2i-ls-06: 0 ≤ i ⊦ (* i) ≤ n→i(n) ↔ i→n(i) ≤ n; from specification intnat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%twols-01: ⊦ m ≠ 1 → (m < 2 ↔ m = 0); from specification nat-basic1%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%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i; from specification int-pot%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs