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Proof step 16 from proof unmodified-clean-eq-keys-subseteq (previous step)

# zbrar0 = BRANCH_SIZE, BRANCH_SIZE = # zbrar1, iast1[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), iast0[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), n < nused, n < BRANCH_SIZE, adr ∈ iast1, adr ∈ iast0, nused ≤ BRANCH_SIZE, ∀ n0. n0 < BRANCH_SIZE → save(zbrar1, 0, nused)[n0] = save(zbrar0, 0, nused)[n0] ⊦ zbrar1[n].key = zbrar0[n].key

User applied rule all left

And obtained the following new goals

Proof step 17:
# zbrar0 = BRANCH_SIZE, BRANCH_SIZE = # zbrar1, iast0[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), iast1[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), n < BRANCH_SIZE, n < nused, adr ∈ iast0, adr ∈ iast1, nused ≤ BRANCH_SIZE, n < BRANCH_SIZE → save(zbrar1, 0, nused)[n] = save(zbrar0, 0, nused)[n] ⊦ zbrar1[n].key = zbrar0[n].key

Rule argument

pos: left 10 terms: n abort: F discard: T computed: T

Additional information

Used terms: n