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Proof step 4 from proof div-mod (
previous step
)
i - i0 = ~(i0 - i), ¬(0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i) ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | i - i0 ↔ i % j = i0 % j)
User applied rule insert equation
And obtained the following new goals
Proof step 5
:
¬(0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i) ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | ~(i0 - i) ↔ i % j = i0 % j)
Rule argument
pos: left 1 flags: discard
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