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Proof step 104 from proof onlymod-shift (previous step)

r1 -- _type ∈ st1, r1 -- _type ∈ st, r1 ∈ st1, r1 ∈ st, ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1, ∀ r, sk0. r ≠ r1 → st[r -- sk0] = st1[r -- sk0] ⊦ r0 = r1, st[r0 -- sk] = st1[r0 -- sk], r1 -- _length ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 6, [QUANTTERMLIST : (r0, sk)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r0, sk)]]