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Proof step 13 from proof mod-square-10 (
previous step
)
i * i = j * 10, 2 | i ⊦ i % 10 = 0
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 14
:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 15
:
i * i = j * 10, ∃ i0. 2 * i0 = i ⊦ i % 10 = 0
Rule argument
spec: inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i, 2] flags: rewrite, current
Additional information
Used terms: i, 2