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Proof step 2 from proof case-20 (previous step)

i < 20, 10 ≤ i ⊦ i = 10, i = 11, i = 12, i = 13, i = 14, i = 15, i = 16, i = 17, i = 18, i = 19

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
1 < i, 0 < i, 10 < i, i < 20 ⊦ i = 10, i = 11, i = 12, i = 13, i = 14, i = 15, i = 16, i = 17, i = 18, i = 19

Rule argument

Additional information

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