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Proof step 55 from proof inv-get-inverse-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
# nσ = ∞, nσ[m1] = nσ[n1], nσ[m1 + 1] = nσ[n1] + 1, n1 ≤ m1, stutf(nσ), ∀ 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: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2