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Proof step 19 from proof two-le-three (previous step)

i ≠ 0 ∧ i ≠ 1, 1 < i ^2, 0 < i ^2, 3 < i ^2, i = ~1, abs(i) < i ^2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
3 < ~1 ^2, 0 < ~1 ^2, 1 < ~1 ^2 ⊦

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%numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1