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Proof step 1 from proof Sep-forall-prove-imp (conclusion)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ n0. n0 < m0 +1 → 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, m0 +1) ⊆ nP1 ⊗_(n, m0 +1)

Rule argument

term: m

Additional information

Used simplifier rules: Sep-base: ⊦ nP ⊗_(n, 0) = emp;%nat freely generated by 0, +1;;% ⊦ P ⊆ P; (s)