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Proof step 5 from proof Sep-forall-reduce-imp (previous step)

((nP0 ⊗_(n, m) * P))(H), ∀ n0. n0 < m → nP0(n + n0) ⊆ nP1(n + n0) ⊦ ((nP1 ⊗_(n, m) * P))(H), nP0 ⊗_(n, m) ⊆ nP1 ⊗_(n, m), ∀ n0. n0 < m → nP0(n + n0) ⊆ nP1(n + n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((nP0 ⊗_(n, m) * P))(H), n0 < m, ∀ n0. n0 < m → nP0(n + n0) ⊆ nP1(n + n0) ⊦ ((nP1 ⊗_(n, m) * P))(H), nP0 ⊗_(n, m) ⊆ nP1 ⊗_(n, m), nP0(n + n0) ⊆ nP1(n + n0)

Rule argument

Additional information

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