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Proof step 15 from proof is-encoded-prefix (previous step)

buf0 + buf1 = buf2 + buf3, n < # buf0, n < # buf2, is-encoded(a, buf0), # buf0 ≤ # buf2 ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b, buf2[n] = buf0[n], is-encoded(a, subarray(buf2, 0, # buf0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n])
Proof step 17:
#(buf0 + buf1) = #(buf2 + buf3), n < # buf2, n < # buf0, is-encoded(a, buf0), # buf0 ≤ # buf2, ∀ n. n < #(buf0 + buf1) → (buf0 + buf1)[n] = (buf2 + buf3)[n] ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b, buf2[n] = buf0[n], is-encoded(a, subarray(buf2, 0, # buf0))

Rule argument

spec: array, inst: buffer name: Extension seq: ⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n]) subst: [buf0, buf] → [buf2 + buf3, buf0 + buf1] flags: <all paths>

Additional information

Used terms: buf2 + buf3, buf0 + buf1