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Proof step 27 from proof equal-minus-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: one-01: ⊦ 1 ^ j = 1;%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%neg-02: ⊦ ~ ~ i = i; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%neg-05: ⊦ ~ i < 0 ↔ 0 < i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: ~ i