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Proof step 35 from proof inv-inverts (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 36:
# nσ = ⌜n1⌝ , nσ[n + 1] = nσ[n] + 1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , n1 = m0 + 1, 1 + n < n1, stutf(nσ) ⊦ ((λ m. max∞(λ n0. n0 < n1 ∧ nσ[n0] = m).n) to nσ[m0] + 1)[nσ[n]] = n, n1 = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n1, m0]

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false);% ⊦ ¬ m < 0; (s)% ⊦ # nσ = ⌜0⌝ ↔ nσ = []; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m0