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Proof step 36 from proof inv-get-inverse (previous step)

stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ nσ[n] + 1 ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = nσ[n] + 1), # nσ = ∞, nσ[n] < nσ[m], n < m, stutf(nσ), ∀ n1. nσ[n1] ≠ nσ[n] + 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
# nσ = ∞, nσ[m0] = nσ[n] + 1, stutf(nσ), n < m, nσ[n] < nσ[m], m0 ≤ m, nσ[m0] ≤ nσ[m], ∀ n1. nσ[n1] ≠ nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat