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Proof step 4 from proof read-write-other-fs (previous step)

∀ fs, adr, nd. fs[adr, nd][ax0] = fs[ax0] ∨ adr ∈ ax0 ⊦ fs[adr, nd][adr0 + ax0] = fs[adr0 + ax0], adr ∈ adr0 + ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ fs, adr, nd. fs[adr, nd][ax0] = fs[ax0] ∨ adr ∈ ax0 ⊦ fs[adr, nd][ax0] = fs[ax0], adr = adr0, adr ∈ ax0

Rule argument

Additional information

Used simplifier rules: read-rec: ⊦ fs[adr ' + ax] = fs[adr] + fs[ax];%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ nd ' + ndx = nd ' + ndx0 ↔ ndx = ndx0; (s)% ⊦ adr ≠ adr0 → fs[adr0, nd][adr] = fs[adr]; (s)% ⊦ adr ∈ adr0 ' + ax ↔ ¬(adr ≠ adr0 ∧ ¬ adr ∈ ax); (ss)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ nd + ndx = nd ' + ndx; (s)% ⊦ adr + ax = adr ' + ax; (s)