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Proof step 75 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 76:
# nσ = ∞, m0 ≤ m, nσ[m0] ≤ nσ[m], stutf(nσ), ∀ n0. nσ[n0] ≠ nσ[m0], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[∞.n + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf