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Proof step 2 from proof read-split (previous step)

fs[ax] = ndx0 + ndx1 ⊦ ∃ ax0, ax1. ax = ax0 + ax1 ∧ fs[ax0] = ndx0 ∧ fs[ax1] = ndx1

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
fs[adr + ax0] = ndx0 + ndx1, ∀ fs, ndx0, ndx1. fs[ax0] = ndx0 + ndx1 → (∃ ax2, ax1. ax0 = ax2 + ax1 ∧ fs[ax2] = ndx0 ∧ fs[ax1] = ndx1) ⊦ ∃ ax2, ax1. adr + ax0 = ax2 + ax1 ∧ fs[ax2] = ndx0 ∧ fs[ax1] = ndx1

Rule argument

term: ax

Additional information

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