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Proof step 4 from proof get-eqex-03 (previous step)

j0 < j, ∀ rk. (rk ∈ st -- r0 ↔ rk ∈ st0 -- r0) ∧ (st -- r0)[rk] = (st0 -- r0)[rk], eqex(st0, st, r0), 0 ≤ j0, 0 < j ⊦ st0[r -- (i + j0) '] = st[r -- (i + j0) '], r = r0

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (r -- (i + j0) ') abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st], [CSIMPSEQ : ⊦ r ≠ r0 → (st -- r)[r0 -- sk0] = st[r0 -- sk0]], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])], [EXTRATERMS : (r -- (i + j0) ')]]