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Proof step 50 from proof eq-leninv (previous step)

m + 1 = nσ[n], # nσ = ∞, stutf(nσ), ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ ⌜m⌝ + ⌜1⌝ = leninv(nσ)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 51:
⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1);nσ[(# nσ).n - 1] = n))
Proof step 52:
stutf(nσ), m + 1 = nσ[n], # nσ = ∞, stutf(nσ), ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ (# nσ = ∞ ⊃ ∃ m0. nσ[m0] = m ∧ (∀ m1. m0 < m1 → nσ[m1] = m + 1);nσ[(# nσ).n - 1] = m)

Rule argument

spec: inst: name: leninv-succ seq: ⊦ stutf(nσ) → ( leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1) ;nσ[(# nσ).n - 1] = n )) subst: [nσ, n] → [nσ, m] flags: rewrite, current

Additional information

Used terms: nσ, m