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Proof step 6 from proof bits2int-nls-01 (previous step)

# bits = m + 1 ∧ # bits ≠ 0, 0 ≤ 2 ^ n→i(m) + bits2int(bits), 2 ^ n→i(n) + bits2int(bits) < 0, # bits ≤ n, bits2int(bits) < 2 ^ n→i(m), 2 ^ n→i(n) < 2 ^ n→i(m) ⊦ bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits = m + 1, 2 ^ n→i(n) + bits2int(bits) < 0, 0 ≤ 2 ^ n→i(m) + bits2int(bits), 2 ^ n→i(n) < 2 ^ n→i(m), bits2int(bits) < 2 ^ n→i(m), m < n ⊦ bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ # bits = 0 ↔ bits = [];