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Proof step 29 from proof is-encoded-prefix (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n])
Proof step 31:
buf0 + buf1 = buf2 + buf3, # buf2 < # buf0, is-encoded(a, buf0), is-encoded(b, buf2) ⊦ # buf0 = # buf2, is-encoded(b, subarray(buf0, 0, # buf2)), # subarray(buf0, 0, # buf2) = # buf2 ∧ (∀ n. n < # subarray(buf0, 0, # buf2) → subarray(buf0, 0, # buf2)[n] = buf2[n])

Rule argument

spec: array, inst: buffer name: Extension seq: ⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n]) subst: [buf0, buf] → [buf2, subarray(buf0, 0, # buf2)] flags: <all paths>

Additional information

Used terms: buf2, subarray(buf0, 0, # buf2)