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Proof step 1 from proof read-split (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
fs[ax] = ndx0 + ndx1 ⊦ ∃ ax0, ax1. ax = ax0 + ax1 ∧ fs[ax0] = ndx0 ∧ fs[ax1] = ndx1

Rule argument

Additional information

Used simplifier rules: read-app: ⊦ fs[ax0 + ax1] = fs[ax0] + fs[ax1];%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ ndx + ndx0 = ndx + ndx1 ↔ ndx0 = ndx1; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]