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Proof step 2 from proof onlymod-other (previous step)

r -- _length ∈ st → st[r -- _length] = st0[r -- _length], r -- _type ∈ st → st[r -- _type] = st0[r -- _type], ∀ r0, sk. r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk], ∀ rk. rk ∈ st ↔ rk ∈ st0 ⊦ st[r0 -- sk] = st0[r0 -- sk], r = r0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (r0, sk)]]

Additional information

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