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Proof step 11 from proof mininv-inverts-inner (previous step)

# nσ = ⌜min(nσ, nσ[n])⌝ , min(nσ, nσ[n]) = m + 1, nσ[m] < nσ[n], n < min(nσ, nσ[n]), mon<(nσ) ⊦ min(nσ, nσ[n]) = n, min(nσ, nσ[n]) = 0, nσ = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# nσ = ⌜m⌝ + ⌜1⌝ , min(nσ, nσ[n]) = m + 1, nσ[m] < nσ[n], mon<(nσ), n ≤ m ⊦ m + 1 = n, nσ = []

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ 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)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)