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Proof step 40 from proof inv-inverts-leq (previous step)

# nσ = ∞, stutf(nσ), nσ[n] ≤ nσ[m], max∞(λ n0. nσ[n0] = nσ[n]).n < n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 41:
# nσ = ∞, stutf(nσ), nσ[n] ≤ nσ[m], max∞(λ n0. nσ[n0] = nσ[n]).n < n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ max∞(λ n0. nσ[n0] = nσ[n]) ≠ ∞
Proof step 54:
max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n1⌝ ∧ max∞(λ n0. nσ[n0] = nσ[n]) ≠ ∞, # nσ = ∞, stutf(nσ), nσ[n] ≤ nσ[m], n1 < n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [max∞(λ n0. nσ[n0] = nσ[n]), m] flags: []

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ[n0] = nσ[n]), m