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Proof step 3 from proof inv-get-inverse-succ-inf (previous step)

# nσ = ∞, ⌜n⌝ < leninv(nσ), stutf(nσ) ⊦ nσ[inv(nσ)[n] + 1] = n + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: inv-get-inverse-succ seq: ⊦ stutf(nσ) ∧ ⌜n⌝ < leninv(nσ) ∧ ⌜inv(nσ)[n]⌝ + ⌜1⌝ < # nσ → nσ[inv(nσ)[n] + 1] = n + 1 subst: [n, nσ] → [n, nσ] flags: <all paths>

Additional information

Used simplifier rules: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ ⌜m⌝ ≠ ∞; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf Used terms: n, nσ