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Proof step 22 from proof isempty-append (previous step)

n < # buf0 + # buf1, ∀ n. n < # buf0 → buf0[n] = empty, ∀ n. n < # buf1 → buf1[n] = empty ⊦ (buf0 + buf1)[n] = empty

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ n < # buf + # buf0 → (buf + buf0)[n] = (n < # buf ⊃ buf[n];buf0[n - # buf])
Proof step 24:
n < # buf0 + # buf1, ∀ n. n < # buf1 → buf1[n] = empty, ∀ n. n < # buf0 → buf0[n] = empty ⊦ (n < # buf0 ⊃ buf0[n];buf1[n - # buf0]) = empty

Rule argument

spec: array-splice, inst: buffer name: append-at seq: ⊦ n < # buf + # buf0 → (buf + buf0)[n] = (n < # buf ⊃ buf[n];buf0[n - # buf]) subst: [n, buf0, buf] → [n, buf1, buf0] flags: <all paths>

Additional information

Used terms: n, buf1, buf0