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Proof step 12 from proof btree-sub-clean-next (previous step)

zh[zbrar[n].child] = mkznode(r1, zbrar0, boolvar1, boolvar2, m0), zh[r] = mkznode(r0, zbrar, false, boolvar0, m), boolvar2, zbrar[n].branch?, n < m, zbrar[n].child ∈ zh, r ∈ zh, ∀ n0. n0 < m ∧ n0 < n ∧ zbrar[n0].branch? ∧ zbrar[n0].child ∈ zh → ¬ zh[zbrar[n0].child].dirty, ∀ n0. n0 < m ∧ ¬ n < n0 ∧ 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 13:
zh[zbrar[n].child] = mkznode(r1, zbrar0, boolvar1, true, m0), zh[r] = mkznode(r0, zbrar, false, boolvar0, m), zbrar[n].branch?, n < m, zbrar[n].child ∈ zh, r ∈ zh, ∀ n0. n0 < m ∧ n0 < n ∧ zbrar[n0].branch? ∧ zbrar[n0].child ∈ zh → ¬ zh[zbrar[n0].child].dirty, ∀ n0. n0 < m ∧ ¬ n < n0 ∧ zbrar[n0].branch? ∧ zbrar[n0].child ∈ zh → ¬ zh[zbrar[n0].child].dirty ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic