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Proof step 5 from proof btree-inv-root-null (previous step)

((btree(null, null, 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, null = null ∨ ¬ zh[null].dirty ⊦ ri = fi

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((btree(null, null, adr, height0, false, fi, nodes(istore0), as0, alladrs0, ok) * true))(zh), ((btree(null, null, adr, height, false, ri, nodes(istore), as, alladrs, ok) * true))(zh), refsize-cons(istore0, as0), refsize-cons(istore, as), adr ∈ istore, as0 ⊆ alladrs0, as0 ⊆ dom nodes(istore0), as ⊆ alladrs, as ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore), alladrs ⊆ dom nodes(istore), as0 ⊆ dom nodes(istore), istore0 ⊑ istore ⊦ ri = fi

Proof step 5 from proof btree-inv-root-null (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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