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Proof step 90 from proof onlymod-shift (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 89:
r1 ∈ st1, r1 ∈ st, r1 -- sk0 ∈ st1, r1 -- sk0 ∈ st, r -- _length ∈ st → (((r = r1 ∧ _length = sk0) ⊃ val; st[r -- _length])) = st1[r -- _length], r -- _type ∈ st → (((r = r1 ∧ _type = sk0) ⊃ val; st[r -- _type])) = st1[r -- _type], ∀ r2, sk1. r2 ≠ r → (((r2 = r1 ∧ sk1 = sk0) ⊃ val; st[r2 -- sk1])) = st1[r2 -- sk1], r0 ≠ r → (((r0 = r1 ∧ sk = sk0) ⊃ val; st[r0 -- sk])) = st1[r0 -- sk], ∀ rk0. rk0 ∈ st ↔ rk0 ∈ st1, sk0 = _length ∨ sk0 = _type ∨ r ≠ r1 ⊦ st1[r1 -- sk0] = val

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]