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Proof step 14 from proof btree-sub-clean-write-branches (previous step)

zh[r] = mkznode(r0, zbrar, false, boolvar0, m), zh[zbrar[n].child] = mkznode(r1, zbrar0, boolvar1, false, m0), zbrar[n, mkzbranch(key, adr, zbrar[n].child)][n0] = mkzbranch(key0, adr0, r2), zbrar[n, mkzbranch(key, adr, zbrar[n].child)][n0].branch?, (r2 = r ⊃ mkznode(r0, zbrar[n, mkzbranch(key, adr, zbrar[n].child)], false, boolvar0, m);zh[r2]).dirty, n < # zbrar, n0 < m, n0 < # zbrar, r ∈ zh, r2 ∈ zh, n0 ≤ n, ∀ n0. n0 < m ∧ n0 < n ∧ zbrar[n0].branch? ∧ zbrar[n0].child ∈ zh → ¬ zh[zbrar[n0].child].dirty ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
zh[r] = mkznode(r0, zbrar, false, boolvar0, m), zh[zbrar[n].child] = mkznode(r1, zbrar0, boolvar1, false, m0), zbrar[n, mkzbranch(key, adr, zbrar[n].child)][n0] = mkzbranch(key0, adr0, r2), (r2 = r ⊃ mkznode(r0, zbrar[n, mkzbranch(key, adr, zbrar[n].child)], false, boolvar0, m);zh[r2]).dirty, n < # zbrar, n0 < m, n0 < # zbrar, r ∈ zh, r2 ∈ zh, n0 ≤ n, ∀ n0. n0 < m ∧ n0 < n ∧ zbrar[n0].branch? ∧ zbrar[n0].child ∈ zh → ¬ zh[zbrar[n0].child].dirty ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ mkzbranch(key, adr, r).branch?; (s)