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Proof step 11 from proof unmodified-clean-eq-keys (previous step)

# save(zbrar1, 0, nused) = # save(zbrar0, 0, nused), # zbrar0 = BRANCH_SIZE, BRANCH_SIZE = # zbrar1, iast[adr] = indexnode(save(zbrar0, 0, nused), bv, nused), n < nused, n < BRANCH_SIZE, adr ∈ iast, nused ≤ BRANCH_SIZE, ∀ n. n < # save(zbrar1, 0, nused) → save(zbrar1, 0, nused)[n] = save(zbrar0, 0, nused)[n] ⊦ zbrar1[n].key = zbrar0[n].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# zbrar0 = BRANCH_SIZE, BRANCH_SIZE = # zbrar1, iast[adr] = indexnode(save(zbrar0, 0, nused), bv, nused), n < BRANCH_SIZE, n < nused, adr ∈ iast, nused ≤ BRANCH_SIZE, ∀ n0. n0 < BRANCH_SIZE → save(zbrar1, 0, nused)[n0] = save(zbrar0, 0, nused)[n0] ⊦ zbrar1[n].key = zbrar0[n].key

Rule argument

Additional information

Used simplifier rules: ⊦ # save(zbrar, m, nused) = BRANCH_SIZE; (s)