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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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