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Proof step 21 from proof int2bits-leneq-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int2bits inst: name: int2bits-len-pos seq: 0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2) subst: [i, j] → [n→i(n) - 2, j] flags: rewrite, current

Additional information

Used simplifier rules: n2n: ⊦ i→n(n→i(n)) = n; from specification intnat%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%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%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs Used terms: n→i(n) - 2, j