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Proof step 5 from proof i2n-divtwo (previous step)

¬ 0 ≤ 2 * i, i→n(2 * i) = m0 * 2 + n0, n0 < 2 ⊦ m0 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i→n(2 * i) = m0 * 2 + n0, n0 < 2, i < 0 ⊦ m0 = i→n(i)

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult