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Proof step 8 from proof read-split (previous step)

fs[ax0] = ndx + ndx1, ∀ 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] = fs[adr] ' + ndx ∧ fs[ax1] = ndx1

User applied rule all left

And obtained the following new goals

Proof step 9:
fs[ax0] = ndx + ndx1, fs[ax0] = ndx + ndx1 → (∃ ax2, ax1. ax0 = ax2 + ax1 ∧ fs[ax2] = ndx ∧ fs[ax1] = ndx1) ⊦ ∃ ax, ax1. adr ' + ax0 = ax + ax1 ∧ fs[ax] = fs[adr] ' + ndx ∧ fs[ax1] = ndx1

Rule argument

pos: left 2 terms: fs, ndx, ndx1 abort: F discard: T computed: T

Additional information

Used terms: fs, ndx, ndx1