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Proof step 23 from proof eq-leninv (previous step)

⌜m⌝ + ⌜1⌝ = leninv(nσ), m + 1 = nσ[n], ⌜n⌝ < # nσ, stutf(nσ) ⊦ m + 1 = 0, # nσ = ∞ ∧ (n < m0 ∧ ⌜m0⌝ < # nσ → nσ[m0] = m + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
⌜m⌝ + ⌜1⌝ = leninv(nσ), m + 1 = nσ[n], ⌜n⌝ < # nσ, stutf(nσ) ⊦ # nσ = ∞ ∧ (n < m0 ∧ ⌜m0⌝ < # nσ → nσ[m0] = nσ[n])

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2