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Proof step 8 from proof int2bits-leneq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%num-i2n-eq: 0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m); from specification intnat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%num-n2i-eq: 0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i); from specification intnat%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0; from specification intnat%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%l-num-04: n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0; from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat%num-n2i-ls-06: 0 ≤ i ⊦ (* i) ≤ n→i(n) ↔ i→n(i) ≤ n; from specification intnat%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