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Proof step 3 from proof onlymod-other (previous step)

(∀ 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]) ⊦ r = r0, st[r0 -- sk] = st0[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
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 ⊦ st[r0 -- sk] = st0[r0 -- sk], r = r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]