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Proof step 2 from proof read-eq-one (previous step)

fs[ax] = nd ' ⊦ ∃ adr. ax = adr ' ∧ fs[adr] = nd

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: ax

Additional information

Used simplifier rules: read-base: ⊦ fs[[]] = [];%read-rec: ⊦ fs[adr ' + ax] = fs[adr] + fs[ax];%read-eq-empty: ⊦ fs[ax] = [] ↔ ax = [];%address-list freely generated by [], +;;% ⊦ adr + ax = adr ' + ax; (s)% ⊦ nd + ndx = nd ' + ndx; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ nd ' = nd0 ' + ndx ↔ nd = nd0 ∧ ndx = []; (ws)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ adr ' = adr0 ' + ax ↔ adr = adr0 ∧ ax = []; (ws)% ⊦ nd ' ≠ []; (s)