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Proof step 28 from proof equal-minus-one (previous step)

i + ~ i = 0, i ^ i0 = ~ i, abs(~ i) = 1, 0 < i0 ⊦ i = ~1, i = 0, i0 = 0, i < 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [~ i] flags: <all paths>

Additional information

Used simplifier rules: one-01: ⊦ 1 ^ j = 1;% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ 0 - i = ~ i; (s)% ⊦ ~ ~ i = i; (s)% ⊦ ~ i + i = 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ i < j → ¬ j < i; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ abs(~ i) = abs(i); (s)% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: ~ i