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Proof step 121 from proof onlymod-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 120:
r1 ∈ st, r1 ∈ st1, r1 -- sk0 ∈ st, r1 -- sk0 ∈ st1, r0 ≠ r1 ∨ sk ≠ sk0, r -- _type ∈ st → st[r -- _type] = (((r = r1 ∧ _type = sk0) ⊃ st[r1 -- _type]; st1[r -- _type])), ∀ r2, sk1. r2 ≠ r → st[r2 -- sk1] = (((r2 = r1 ∧ sk1 = sk0) ⊃ st[r1 -- sk0]; st1[r2 -- sk1])), ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1, sk0 = _type ∨ r ≠ r1 → st1[r1 -- sk0] = val ⊦ st[r0 -- sk] = st1[r0 -- sk], r0 = r, r -- _length ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]