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Proof step 4 from proof elim-fromto-inf (previous step)

x0 = nf to n ∧ nf0 = nf from n ↔ ⌜nf⌝ = ⌜x0⌝ + ⌜nf0⌝ ∧ # x0 = n ⊦ x0 = nf to n ∧ nf0 = nf from n ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # x0 = n

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: append-fininf seq: ⊦ ⌜x⌝ + ⌜nf⌝ = ⌜λ n. (n < # x ⊃ x[n];nf(n - # x))⌝ subst: [nf, x] → [nf0, x0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)% ⊦ ⌜nf⌝ .f = nf; (s) Used terms: nf0, x0