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Proof step 22 from proof is-encoded-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ is-encoded(a, buf) → (is-encoded(b, buf) ↔ a = b)
Proof step 24:
buf0 + buf1 = buf2 + buf3, # buf2 < # buf0, is-encoded(a, buf0), is-encoded(b, subarray(buf0, 0, # buf2)) → b = a ∧ # buf2 = # buf0 ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b

Rule argument

Additional information

Used simplifier rules: is-encoded-unique: ⊦ is-encoded(a, buf) → (is-encoded(b, buf) ↔ a = b);%# buf = n ⊦ subarray(buf, 0, n) = buf; (s)% ⊦ m < n → m ≤ n; (s)