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Proof step 8 from proof at-write-at (previous step)

1 + # ax0 = # ndx ⊦ fs[adr ' + ax0, ndx][adr] = ndx[0], dups(ax0), adr ∈ ax0

User applied rule constructor cut

And closed the goal!

Rule argument

var: ndx

Additional information

Used simplifier rules: node-list freely generated by [], +;;% ⊦ nd + ndx = nd ' + ndx; (s)% ⊦ #(nd ' + ndx) = (# ndx)+1; (s)% ⊦ m +1 = m + 1; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ fs[adr ' + ax, nd ' + ndx] = fs[adr, nd][ax, ndx]; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ # ax = # ndx ∧ ¬ adr ∈ ax → fs[ax, ndx][adr] = fs[adr]; (s)% ⊦ fs[adr, nd][adr] = nd; (s)% ⊦ (nd ' + ndx)[0] = nd; (s)% ⊦ # [] = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)