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Proof step 5 from proof onlymod-sym-01 (previous step)

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

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
rk ∈ st0 ↔ rk ∈ st, r -- _length ∈ st0 ↔ r -- _length ∈ st, r -- _type ∈ st0 ↔ r -- _type ∈ st, r -- _length ∈ st → st0[r -- _length] = st[r -- _length], r -- _type ∈ st → st0[r -- _type] = st[r -- _type], r0 ≠ r → st0[r0 -- sk] = st[r0 -- sk], r ≠ r → st0[r -- _type] = st[r -- _type], ∀ r0, sk. r0 ≠ r → st0[r0 -- sk] = st[r0 -- sk], ∀ rk. rk ∈ st0 ↔ rk ∈ st ⊦ (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])

Rule argument

[EXRARG : pos: left 7, [QUANTTERMLIST : (r, _type)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, _type)]]