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Proof step 11 from proof append-assoc (previous step)

⊦ (⌜x⌝ + ⌜x0⌝ ) + ⌜nf⌝ = ⌜x⌝ + ⌜λ n. (n < # x0 ⊃ x0[n];nf(n - # x0))⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ ⌜x⌝ + ⌜nf⌝ = ⌜λ n. (n < # x ⊃ x[n];nf(n - # x))⌝
Proof step 13:
⊦ (⌜x⌝ + ⌜x0⌝ ) + ⌜nf⌝ = ⌜λ n. (n < # x ⊃ x[n];(n - # x < # x0 ⊃ x0[n - # x];nf(n - # x - # x0)))⌝

Rule argument

spec: stream inst: name: append-fininf seq: ⊦ ⌜x⌝ + ⌜nf⌝ = ⌜λ n. (n < # x ⊃ x[n];nf(n - # x))⌝ subst: [nf, x] → [λ n. (n < # x0 ⊃ x0[n];nf(n - # x0)), x] flags: rewrite, current

Additional information

Used terms: λ n. (n < # x0 ⊃ x0[n];nf(n - # x0)), x