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Proof step 55 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, nσ[n0] = nσ[n], ⌜n0⌝ < # nσ, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦ (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ )[nσ[n]] = n, n0 = n

User applied rule all left

And obtained the following new goals

Proof step 56:
nσ[n + 1] = nσ[n] + 1, nσ[n0] = nσ[n], ⌜n0⌝ < # nσ, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n] → ¬ n0 < n0, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦ (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to leninv(nσ).n⌝ )[nσ[n]] = n, n0 = n

Rule argument

pos: left 6 terms: n0 abort: F discard: F computed: T

Additional information

Used terms: n0