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Proof step 15 from proof int2bytes-minus-one (previous step)

¬ 0 < i, i2b(i) = i2b(~1), # int2bits(i) ≤ 8 ⊦ i = ~1, ¬ i < ~128 ∧ ¬ 127 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
i2b(i) = i2b(~1), i < 0, i < ~128, # int2bits(i) ≤ 8 ⊦ i = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;