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Proof step 4 from proof in-write (previous step)

# ax0 + 1 = # ndx, ndx = nd + ndx0, ∀ ndx, adr, fs. # ax0 = # ndx → (adr ∈ fs[ax0, ndx] ↔ adr ∈ fs ∨ adr ∈ ax0) ⊦ adr ∈ fs[adr0 ' + ax0, ndx] ↔ adr ∈ fs ∨ adr = adr0 ∨ adr ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# ax0 = # ndx0, ∀ ndx, adr, fs. # ndx0 = # ndx → (adr ∈ fs[ax0, ndx] ↔ adr ∈ fs ∨ adr ∈ ax0) ⊦ adr ∈ fs[adr0, nd][ax0, ndx0] ↔ adr ∈ fs ∨ adr = adr0 ∨ adr ∈ ax0

Rule argument

Additional information

Used simplifier rules: write-rec: ⊦ fs[adr ' + ax, nd ' + ndx] = fs[adr, nd][ax, ndx];%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ m +1 = m + 1; (s)% ⊦ #(nd ' + ndx) = (# ndx)+1; (s)% ⊦ nd + ndx = nd ' + ndx; (s)