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Proof step 4 from proof Sep-split (previous step)

⊦ nP ⊗_(n, n0 + m0) = nP ⊗_(n, m0) * nP ⊗_(n + m0, n0)

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦ nP ⊗_(n, n0 + m +1) = nP ⊗_(n, m +1) * nP ⊗_(n + m +1, n0)

Rule argument

indvar: m0

Additional information

Used simplifier rules: Sep-base: ⊦ nP ⊗_(n, 0) = emp;%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ emp * P = P; (s)%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep