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Proof step 4 from proof countrefs-le-02 (previous step)

0 < i, countrefs(ints) * i < countrefs(ints) ⊦ ints = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 0 < i → (i * j < i ↔ j < 1)
Proof step 3:
⊦ ¬ countrefs(ints) < 0

Rule argument

[speclemmaarg: ,0 < i → (i * j < i ↔ j < 1) ,[j, i] → [i, countrefs(ints)] ,F ,int-mult , ,ls-same ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : i + 1 = i0 ⊦ j < ((* i0)) → ¬ ((* i)) < j], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPFORWARD : ⊦ i ≤ j ∧ j < j0 → i < j0], [CSIMPFORWARD : ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0], [CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPSEQ : ⊦ 0 < i → (i * j < 0 ↔ j < 0)], [CSIMPSEQ : ⊦ i * j * j0 = (i * j) * j0], [CSIMPSEQ : ⊦ i * j = j * i])], [EXTRATERMS : (i, countrefs(ints))]]