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Proof step 130 from proof onlymod-shift (previous step)

st[r -- _length] = st1[r -- _length], r1 -- sk0 ∈ st1, r1 -- sk0 ∈ st, r1 ∈ st1, r1 ∈ st, r ≠ r1 → st1[r1 -- sk0] = val, ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1, ∀ r2, sk1. r2 ≠ r → st[r2 -- sk1] = (((r2 = r1 ∧ sk1 = sk0) ⊃ st[r1 -- sk0]; st1[r2 -- sk1])) ⊦ r0 = r1, _length = sk0, r0 = r, st[r0 -- sk] = st1[r0 -- sk], r -- _type ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 8, [QUANTTERMLIST : (r0, sk)]]

Additional information

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