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Proof step 9 from proof less-leninv (previous step)

m = m0 +1, nσ[n + 1] = nσ[n] + 1, # nσ = ∞, ⌜m⌝ = leninv(nσ), stutf(nσ), m ≤ nσ[n] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⌜m0⌝ + ⌜1⌝ = leninv(nσ), # nσ = ∞, nσ[n + 1] = nσ[n] + 1, m0 < nσ[n], stutf(nσ) ⊦

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%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ: ⊦ m +1 = m + 1; from specification nat-basic2