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Proof step 64 from proof inv-get-inverse-last (previous step)

# nσ = ⌜n⌝ , nσ[m1] = nσ[n1], n = m1 + 1, 1 + m + n1 < n, n1 < n, stutf(nσ), ∀ m0. m0 < n ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
# nσ = ⌜m1⌝ + ⌜1⌝ , nσ[m1] = nσ[n1], m + n1 < m1, stutf(nσ), n1 ≤ m1, ∀ m0. ¬ m1 < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)