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Proof step 10 from proof Sep-split (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
nP ⊗_(n + 1, n0 + m) = nP ⊗_(n + 1, m) * nP ⊗_((n + 1) + m, n0) ⊦ nP(n) * nP ⊗_(n + 1, m + n0) = (nP(n) * nP ⊗_(n + 1, m)) * nP ⊗_(n + m + 1, n0)

Rule argument

pos: left 1 terms: nP, n + 1, n0 abort: F discard: T computed: F

Additional information

Used terms: nP, n + 1, n0