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Proof step 161 from proof onlymod-shift (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 139:
r = r1 ∧ _type = sk0, st[r -- _type] = st[r1 -- _type], r ≠ r1 ∨ _length ≠ sk0, st[r -- _length] = st1[r -- _length], r0 ≠ r1 ∨ sk ≠ sk0, r1 ∈ st, r1 ∈ st1, r1 -- sk0 ∈ st, r1 -- sk0 ∈ st1, sk0 = _length ∨ sk0 = _type ∨ r ≠ r1 → st1[r1 -- sk0] = val, ∀ r2, sk1. r2 ≠ r → st[r2 -- sk1] = (((r2 = r1 ∧ sk1 = sk0) ⊃ st[r1 -- sk0]; st1[r2 -- sk1])), ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1 ⊦ st[r0 -- sk] = st1[r0 -- sk], r0 = r
Proof step 160:
r ≠ r1 ∨ _type ≠ sk0, st[r -- _type] = st1[r -- _type], r ≠ r1 ∨ _length ≠ sk0, st[r -- _length] = st1[r -- _length], r0 ≠ r1 ∨ sk ≠ sk0, r1 ∈ st, r1 ∈ st1, r1 -- sk0 ∈ st, r1 -- sk0 ∈ st1, sk0 = _length ∨ sk0 = _type ∨ r ≠ r1 → st1[r1 -- sk0] = val, ∀ r2, sk1. r2 ≠ r → st[r2 -- sk1] = (((r2 = r1 ∧ sk1 = sk0) ⊃ st[r1 -- sk0]; st1[r2 -- sk1])), ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1 ⊦ st[r0 -- sk] = st1[r0 -- sk], r0 = r

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]