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Proof step 59 from proof inv-get-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
# nσ = ∞, stutf(nσ), nσ[n0] ≤ nσ[m], m < n0, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1