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Proof step 7 from proof btree-inv-root-null-or-clean (previous step)

zh[r] = mkznode(r0, zbrar, boolvar, boolvar0, m), ((btree(null, r, adr, height, false, ri, nodes(istore), as, alladrs, ok) * true))(zh), ((btree(null, null, adr, height0, false, fi, nodes(istore0), as0, alladrs0, ok) * true))(zh), refsize-cons(istore, as), refsize-cons(istore0, as0), adr ∈ istore, as0 ⊆ dom nodes(istore), alladrs ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore), as ⊆ dom nodes(istore), as ⊆ alladrs, as0 ⊆ dom nodes(istore0), as0 ⊆ alladrs0, istore0 ⊑ istore, r = null ∨ ¬ boolvar0, r = null ∨ ¬ zh[r].dirty ⊦ ri = fi

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
zh[r] = mkznode(r0, zbrar, boolvar, boolvar0, m), ((btree(null, r, adr, height, false, ri, nodes(istore), as, alladrs, ok) * true))(zh), ((btree(null, null, adr, height0, false, fi, nodes(istore0), as0, alladrs0, ok) * true))(zh), refsize-cons(istore, as), refsize-cons(istore0, as0), adr ∈ istore, as0 ⊆ dom nodes(istore), alladrs ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore), as ⊆ dom nodes(istore), as ⊆ alladrs, as0 ⊆ dom nodes(istore0), as0 ⊆ alladrs0, istore0 ⊑ istore, r = null ∨ ¬ boolvar0 ⊦ ri = fi

Rule argument

Additional information

Used simplifier rules: ⊦ mkznode(r, zbrar, boolvar, boolvar0, m).dirty ↔ boolvar0; (s)