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Proof step 10 from proof mininv-inverts-inner-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# nσ = ⌜m⌝ + ⌜1⌝ , min(nσ, nσ[n] + 1) = m + 1, nσ[m] ≤ nσ[n], mon<(nσ), n < m ⊦ m = 0, m = n, nσ = []

Rule argument

Additional information

Used simplifier rules: neq-01: ⊦ m ≠ n → (m < n + 1 ↔ m < n); from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1