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Proof step 3 from proof at-init (previous step)

is_fskey(sk), is_fskey(sk) ⊦ (addarray(r, vals, j, st)[r -- sk, val][r -- i ']) = (addarray(r, vals, j, st)[r -- i '])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
is_fskey(sk) ⊦ ((i ' = sk ⊃ val; addarray(r, vals, j, st)[r -- i '])) = (addarray(r, vals, j, st)[r -- i '])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))])]]