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Proof step 2 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)

User applied rule cut formula

And obtained the following new goals

Proof step 3:
((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)

Rule argument

fma: nP0 ⊗_(n, m) ⊆ nP1 ⊗_(n, m)

Additional information

Used simplifier rules: sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep% ⊦ P0 ⊆ P1 ∧ ((P * P0))(H) → ((P * P1))(H); (s) Used formulas: nP0 ⊗_(n, m) ⊆ nP1 ⊗_(n, m)