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Proof step 10 from proof write-shift (
previous step
)
# ax0 = # ndx0, ∀ ndx, fs, adr, nd. # ndx0 = # ndx → fs[adr, nd][ax0, ndx] = fs[ax0, ndx][adr, nd] ∨ adr ∈ ax0 ⊦ fs[adr0, nd0][adr, nd][ax0, ndx0] = fs[adr0, nd0][ax0, ndx0][adr, nd], adr0 = adr, adr ∈ ax0
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: ndx0, fs[adr0, nd0], adr, nd
Additional information
Used terms: ndx0, fs[adr0, nd0], adr, nd