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

n0 < m0, ∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0) ⊦ nP0(n) * nP0 ⊗_(n + 1, m0) = nP1(n) * nP1 ⊗_(n + 1, m0), nP0((n + 1) + n0) = nP1((n + 1) + n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 < m0, ∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0) ⊦ nP0(n) * nP0 ⊗_(n + 1, m0) = nP1(n) * nP1 ⊗_(n + 1, m0), nP0((n + 1) + n0) = nP1((n + 1) + n0)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic