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Proof step 3 from proof sep-forall-reduce (previous step)

((nP ⊗_(n, m0 +1) * P0))(zh), ∀ n0, P. n0 < m0 +1 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * P))(zh), ∀ nP, n, P0, zh, nP2. ((nP ⊗_(n, m0) * P0))(zh) ∧ (∀ n0, P. n0 < m0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * P))(zh)) → ((nP2 ⊗_(n, m0) * P0))(zh) ⊦ ((nP2 ⊗_(n, m0 +1) * P0))(zh)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
((nP ⊗_(n, m0 + 1) * P0))(zh), ∀ nP, n, P0, zh, nP2. ((nP ⊗_(n, m0) * P0))(zh) ∧ (∀ n0, P. n0 < m0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * P))(zh)) → ((nP2 ⊗_(n, m0) * P0))(zh), ∀ n0, P. ¬ m0 < n0 ∧ ((nP(n + n0) * P))(zh) → ((nP2(n + n0) * P))(zh) ⊦ ((nP2 ⊗_(n, m0 + 1) * P0))(zh)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)