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Proof step 26 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), n ≤ m, m < n + 1 → nσ[n + 1] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , # nσ = ∞, nσ[n + 1] = nσ[n] + 1, n ≤ m, stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ¬ n < m → nσ[n] = nσ[m] ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2