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Proof step 21 from proof onlymod-shift (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: left 7, ()]

Additional information

[[CONCRETESIMPRULES : ()]]