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

∀ nP, n. nP ⊗_(n, m0) = (λ n0. ⌜n0 < n + m0⌝ → nP(n0))⊗_(n, m0) ⊦ nP ⊗_(n, m0 +1) = (λ n0. ⌜n0 < n + m0 +1⌝ → nP(n0))⊗_(n, m0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ nP, n. nP ⊗_(n, m0) = (λ n0. ⌜n0 < n + m0⌝ → nP(n0))⊗_(n, m0) ⊦ nP ⊗_(n, m0 + 1) = (λ n0. ⌜¬ n + m0 < n0⌝ → nP(n0))⊗_(n, m0 + 1)

Rule argument

Additional information

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