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Proof step 19 from proof unmodified-clean-eq-keys-subseteq (previous step)

# zbrar0 = BRANCH_SIZE, BRANCH_SIZE = # zbrar1, iast0[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), iast1[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), mkzbranch(key', adr0, rp) = zbrar0[n], save(zbrar1[n]) = save(zbrar0[n]), n < BRANCH_SIZE, n < nused, adr ∈ iast0, adr ∈ iast1, nused ≤ BRANCH_SIZE ⊦ zbrar1[n].key = zbrar0[n].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# zbrar0 = BRANCH_SIZE, BRANCH_SIZE = # zbrar1, iast0[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), iast1[adr] = indexnode(save(zbrar1, 0, nused), bv, nused), mkzbranch(key', adr0, rp) = zbrar0[n], save(zbrar1[n]) = mkbranch(key', adr0), n < BRANCH_SIZE, n < nused, adr ∈ iast0, adr ∈ iast1, nused ≤ BRANCH_SIZE ⊦ zbrar1[n].key = key'

Rule argument

Additional information

Used simplifier rules: ⊦ mkzbranch(key, adr, r).key = key; (s)% ⊦ save(mkzbranch(key, adr, r)) = mkbranch(key, adr); (s)