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Proof step 50 from proof inv-get-inverse-last (previous step)

nσ[m1] = nσ[n1], # nσ = ∞, stutf(nσ), m1 < n1 → nσ[n1] = nσ[n1] + 1, ∀ m0. m1 < m0 → nσ[m0] = nσ[n1] + 1, ∀ m0. nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
nσ[m1] = nσ[n1], # nσ = ∞, stutf(nσ), n1 ≤ m1, ∀ m0. m1 < m0 → nσ[m0] = nσ[n1] + 1, ∀ m0. nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2