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Proof step 2 from proof unmodified-clean-leaf-subseteq (previous step)

unmodified(adr, mkznode(rp0, zbrar0, bv0, false, nused0), iast0), unmodified(adr, mkznode(rp1, zbrar1, bv1, false, nused1), iast1), iast0 ⊆ iast1 ⊦ (bv0 ↔ bv1) ∧ nused0 = nused1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
iast1[adr] = save(mkznode(rp1, zbrar1, bv1, false, nused1)), iast0[adr] = save(mkznode(rp0, zbrar0, bv0, false, nused0)), adr ∈ iast1, adr ∈ iast0, iast0 ⊆ iast1 ⊦ (bv0 ↔ bv1) ∧ nused0 = nused1

Rule argument

Additional information

Used simplifier rules: unmodified: ⊦ unmodified(adr, znd, iast) ↔ ¬ znd.dirty → adr ∈ iast ∧ iast[adr] = save(znd);% ⊦ mkznode(r, zbrar, boolvar, boolvar0, m).dirty ↔ boolvar0; (s)