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Proof step 167 from proof onlymod-shift (previous step)

st[r -- _type] = (((r = r1 ∧ _type = sk0) ⊃ st[r1 -- _type]; st1[r -- _type])), r -- _type ∈ st, r1 -- sk0 ∈ st1, r1 -- sk0 ∈ st, r1 ∈ st1, r1 ∈ st, r ∈ st, r -- _length ∈ st → st[r -- _length] = (((r = r1 ∧ _length = sk0) ⊃ st[r1 -- _length]; st1[r -- _length])), sk0 = _length ∨ sk0 = _type ∨ 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])) ⊦ (((r = r1 ∧ _type = sk0) ⊃ val; st[r -- _type])) = st1[r -- _type]

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ _type ≠ _length])]]