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Proof step 4 from proof btree-ri-unique (previous step)

rp0 = null, ((btree(rp, rp0, adr, height, dirty, ri0, iast, as0, alladrs0, ok) * P0))(zh), ((btree(rp, rp0, adr, height, dirty, ri1, iast, as1, alladrs1, ok) * P1))(zh), as0 ⊆ alladrs0, as0 ⊆ dom iast, as1 ⊆ alladrs1, as1 ⊆ dom iast ⊦ ri0 = ri1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
((btree(rp, null, adr, height, dirty, ri1, iast, as1, alladrs1, ok) * P1))(zh), ((btree(rp, null, adr, height, dirty, ri0, iast, as0, alladrs0, ok) * P0))(zh), as1 ⊆ dom iast, as1 ⊆ alladrs1, as0 ⊆ dom iast, as0 ⊆ alladrs0 ⊦ ri0 = ri1

Proof step 4 from proof btree-ri-unique (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information