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Proof step 72 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
# nσ = ∞, m + 1 < n0, stutf(nσ), nσ[n0] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[∞.n + 1] = nσ[n0] + 1

Rule argument

Additional information

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