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Proof step 3 from proof le-plus (previous step)

0 < j, 0 ≤ i, abs(i + j) ≤ i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ i → abs(i) = i

Rule argument

spec: int-abs name: absle seq: ⊦ 0 ≤ i → abs(i) = i subst: [i] → [i + j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i + j ≤ i ↔ j ≤ 0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ j < 0 → abs(j) = 0 - j; (s)% ⊦ 0 - i = ~ i; (s)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ i < j → ¬ j < i; (s) Used terms: i + j