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Proof step 12 from proof onlymod-del (previous step)

rk ∈ st, (∀ rk. rk ∈ st ↔ rk ∈ st0) ∧ (∀ r0, sk. r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk]) ∧ (r -- _type ∈ st → st[r -- _type] = st0[r -- _type]) ∧ (r -- _length ∈ st → st[r -- _length] = st0[r -- _length]) ⊦ (rk0 ≠ rk ∧ rk0 ∈ st → rk0 ∈ st0) ∧ (r0 ≠ r → (st -- rk)[r0 -- sk] = (st0 -- rk)[r0 -- sk]) ∧ (r -- _type ≠ rk ∧ r -- _type ∈ st → st[r -- _type] = st0[r -- _type]) ∧ (r -- _length ≠ rk ∧ r -- _length ∈ st → st[r -- _length] = st0[r -- _length])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
rk ∈ st, r -- _length ∈ st → st[r -- _length] = st0[r -- _length], r -- _type ∈ st → st[r -- _type] = st0[r -- _type], ∀ r0, sk. r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk], ∀ rk. rk ∈ st ↔ rk ∈ st0 ⊦ (rk0 ≠ rk ∧ rk0 ∈ st → rk0 ∈ st0) ∧ (r0 ≠ r → (st -- rk)[r0 -- sk] = (st0 -- rk)[r0 -- sk]) ∧ (r -- _type ≠ rk ∧ r -- _type ∈ st → st[r -- _type] = st0[r -- _type]) ∧ (r -- _length ≠ rk ∧ r -- _length ∈ st → st[r -- _length] = st0[r -- _length])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]