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Proof step 4 from proof mon-eq (previous step)

mon<(nσ) ∧ n < n0 ∧ ⌜n0⌝ < # nσ → nσ[n] < nσ[n0], nσ[n] = nσ[n0], mon<(nσ), ⌜n⌝ < # nσ, ⌜n0⌝ < # nσ ⊦ n = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
nσ[n] = nσ[n0], ⌜n0⌝ < # nσ, ⌜n⌝ < # nσ, mon<(nσ), n0 < n ⊦ n = n0, n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1