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Proof step 3 from proof monls-notsmall (previous step)

nσ[0] = 0, ⌜0 + 1⌝ < # nσ, mon<(nσ) ⊦ 0 < nσ[0 + 1] ↔ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
nσ[0] = 0, nσ[1] = 0, mon<(nσ) ⊦ # nσ = ⌜1⌝ , nσ = []

Rule argument

Additional information

Used simplifier rules: ⊦ # nσ = ⌜0⌝ ↔ nσ = []; (s)% ⊦ ⌜1⌝ < n∞ ↔ n∞ ≠ ⌜0⌝ ∧ n∞ ≠ ⌜1⌝ ; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic