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Proof step 6 from proof int2bits-leneq-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 0 < i ∧ 0 ≤ j → (i < 2 ^ j ↔ # int2bits(i) ≤ i→n(j) + 1)
Proof step 8:
0 < j ∧ 0 ≤ n→i(# int2bits(j)) - 2, 2 < # int2bits(j), 0 < j, # int2bits(j) ≤ i→n(n→i(# int2bits(j)) - 2) + 1 ⊦ j = 0

Rule argument

spec: int2bits inst: name: int2bits-lslen-01 seq: ⊦ 0 < i ∧ 0 ≤ j → (i < 2 ^ j ↔ # int2bits(i) ≤ i→n(j) + 1) subst: [j, i] → [n→i(# int2bits(j)) - 2, j] flags: rewrite, current

Additional information

Used simplifier rules: 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%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: n→i(# int2bits(j)) - 2, j