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Proof step 3 from proof is-encoded-pad-unique (previous step)

is-encoded-pad(ns0, n, buf), is-encoded-pad(ns1, n, buf) ⊦ ns0 = ns1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ is-encoded-pad(ns, n, buf) ↔ n = # ns ∧ (∃ buf0, buf1. is-aligned(# buf, EB_PAGE_SIZE) ∧ buf = buf0 + buf1 ∧ is-encoded(ns, buf0))
Proof step 5:
n = # ns0, is-encoded-pad(ns1, n, buf), ∃ buf0, buf1. is-aligned(# buf, EB_PAGE_SIZE) ∧ buf = buf0 + buf1 ∧ is-encoded(ns0, buf0) ⊦ ns0 = ns1

Rule argument

spec: <toplevel> name: is-encoded-pad seq: ⊦ is-encoded-pad(ns, n, buf) ↔ n = # ns ∧ (∃ buf0, buf1. is-aligned(# buf, EB_PAGE_SIZE) ∧ buf = buf0 + buf1 ∧ is-encoded(ns, buf0)) subst: [buf, n, ns] → [buf, n, ns0] flags: <all paths>

Additional information

Used terms: buf, n, ns0