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Proof step 9 from proof onlymod-shift (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 5, [QUANTTERMLIST : (r1, sk0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r1, sk0)]]