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Proof step 61 from proof inv-inverts-leq (previous step)

# nσ = ⌜n1⌝ , ⌜n⌝ < # nσ, stutf(nσ), max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]).n < n, nσ[n] ≤ nσ[n1 - 1] ⊦ # nσ = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
# nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n < n1, n < n1, stutf(nσ), max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n < n, nσ[n] ≤ nσ[n1 - 1], max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]).n + 1 < n1 ⊦

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic