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Proof step 32 from proof is-encoded-prefix (previous step)

buf0 + buf1 = buf2 + buf3, n < # buf0, n < # buf2, # buf2 < # buf0, n + 1 < # buf0, is-encoded(a, buf0), is-encoded(b, buf2) ⊦ # buf0 = # buf2, buf0[n] = buf2[n], is-encoded(b, subarray(buf0, 0, # buf2))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 33:
⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n])
Proof step 34:
#(buf0 + buf1) = #(buf2 + buf3), n + 1 < # buf0, # buf2 < # buf0, n < # buf2, n < # buf0, is-encoded(b, buf2), is-encoded(a, buf0), ∀ n. n < #(buf0 + buf1) → (buf0 + buf1)[n] = (buf2 + buf3)[n] ⊦ # buf0 = # buf2, buf0[n] = buf2[n], is-encoded(b, subarray(buf0, 0, # buf2))

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