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Proof step 6 from proof read-split (previous step)

fs[adr] ' + fs[ax0] = ndx0 + ndx1, fs[adr] ∈ ndx0, ∀ fs, ndx0, ndx1. fs[ax0] = ndx0 + ndx1 → (∃ ax2, ax1. ax0 = ax2 + ax1 ∧ fs[ax2] = ndx0 ∧ fs[ax1] = ndx1) ⊦ ∃ ax, ax1. adr ' + ax0 = ax + ax1 ∧ fs[ax] = ndx0 ∧ fs[ax1] = ndx1

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
fs[adr] ' + fs[ax0] = ndx0 + ndx1, ndx0 = nd + ndx, fs[adr] ∈ ndx0, ∀ fs, ndx0, ndx1. fs[ax0] = ndx0 + ndx1 → (∃ ax2, ax1. ax0 = ax2 + ax1 ∧ fs[ax2] = ndx0 ∧ fs[ax1] = ndx1) ⊦ ∃ ax, ax1. adr ' + ax0 = ax + ax1 ∧ fs[ax] = ndx0 ∧ fs[ax1] = ndx1

Rule argument

var: ndx0

Additional information

Used simplifier rules: node-list freely generated by [], +;;% ⊦ [] + ndx = ndx; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ ¬ nd ∈ []; (s)