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Proof step 7 from proof Sep-forall-prove-imp (previous step)

∀ 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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ P0 ⊆ P1 ∧ P2 ⊆ P3 → P0 * P2 ⊆ P1 * P3
Proof step 9:
∀ 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) ⊆ nP1(n) ∧ nP0 ⊗_(n + 1, m0) ⊆ nP1 ⊗_(n + 1, m0)

Rule argument

spec: heap-sep name: sep-inj2-imp seq: ⊦ P0 ⊆ P1 ∧ P2 ⊆ P3 → P0 * P2 ⊆ P1 * P3 subst: [P3, P1, P2, P0] → [nP1 ⊗_(n + 1, m0), nP1(n), nP0 ⊗_(n + 1, m0), nP0(n)] flags: <all paths>

Additional information

Used terms: nP1 ⊗_(n + 1, m0), nP1(n), nP0 ⊗_(n + 1, m0), nP0(n)