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Proof step 17 from proof int2bits-leneq-neg (previous step)

n→i(# int2bits(j)) = i0 + 2, 2 < # int2bits(j), # int2bits(j) ≤ i→n(i0) + 1, j < 0, j < ~1 ⊦ # int2bits(j) = 1, # int2bits(j) = 3, # int2bits(j) = 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
n→i(# int2bits(j)) = i0 + 2, j < ~1, j < 0, # int2bits(j) ≤ i→n(i0) + 1, 2 < # int2bits(j) ⊦ # int2bits(j) = 2, # int2bits(j) = 3, # int2bits(j) = 1, i→n(i0) = 1, i0 = 0

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%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