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Proof step 136 from proof onlymod-shift (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 129:
r ≠ r1, st[r -- _length] = st1[r -- _length], r1 ∈ st, r1 ∈ st1, r1 -- sk0 ∈ st, r1 -- sk0 ∈ st1, r0 ≠ r1 ∨ sk ≠ sk0, ∀ r2, sk1. r2 ≠ r → st[r2 -- sk1] = (((r2 = r1 ∧ sk1 = sk0) ⊃ st[r1 -- sk0]; st1[r2 -- sk1])), ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1, r ≠ r1 ∨ sk0 = _length → st1[r1 -- sk0] = val ⊦ st[r0 -- sk] = st1[r0 -- sk], r0 = r, r -- _type ∈ st
Proof step 135:
_length ≠ sk0, st[r -- _length] = st1[r -- _length], r1 ∈ st, r1 ∈ st1, r1 -- sk0 ∈ st, r1 -- sk0 ∈ st1, r0 ≠ r1 ∨ sk ≠ sk0, ∀ r2, sk1. r2 ≠ r → st[r2 -- sk1] = (((r2 = r1 ∧ sk1 = sk0) ⊃ st[r1 -- sk0]; st1[r2 -- sk1])), ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1, r ≠ r1 ∨ sk0 = _length → st1[r1 -- sk0] = val ⊦ st[r0 -- sk] = st1[r0 -- sk], r0 = r, r -- _type ∈ st

Rule argument

[CASEDARG : pos: left 6, ()]

Additional information

[[CONCRETESIMPRULES : ()]]