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Proof step 2 from proof Sep-forall (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ 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, m0 + 1) = nP1 ⊗_(n, m0 + 1)

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)