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Proof step 1 from proof case-30 (conclusion)

20 ≤ i, i < 30 ⊦ i = 20 ∨ i = 21 ∨ i = 22 ∨ i = 23 ∨ i = 24 ∨ i = 25 ∨ i = 26 ∨ i = 27 ∨ i = 28 ∨ i = 29

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
1 < i, 0 < i, 20 < i, i < 30 ⊦ i = 29, i = 28, i = 27, i = 26, i = 25, i = 24, i = 23, i = 22, i = 21, i = 20

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs